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1l5bb8fiy | maths | probability | binomial-distribution | <p>In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability $${3 \over 4}$$ and the remaining 6 questions correctly with probability $${1 \over 4}$$. If the probability that the student guesses the answers of exactly 8 questions corre... | [] | null | 479 | <p>Student guesses only two wrong. So there are three possibilities.</p>
<p>(i) Student guesses both wrong from 1<sup>st</sup> section</p>
<p>(ii) Student guesses both wrong from 2<sup>nd</sup> section</p>
<p>(iii) Student guesses two wrong one from each section</p>
<p>Required probabilities</p>
<p>$$ = {}^4{C_2}{\left... | integer | jee-main-2022-online-24th-june-evening-shift | 7,571 |
1l5c1q77f | maths | probability | binomial-distribution | <p>If a random variable X follows the Binomial distribution B(33, p) such that <br/><br/>$$3P(X = 0) = P(X = 1)$$, then the value of $${{P(X = 15)} \over {P(X = 18)}} - {{P(X = 16)} \over {P(X = 17)}}$$ is equal to :</p> | [{"identifier": "A", "content": "1320"}, {"identifier": "B", "content": "1088"}, {"identifier": "C", "content": "$${{120} \\over {1331}}$$"}, {"identifier": "D", "content": "$${{1088} \\over {1089}}$$"}] | ["A"] | null | $3 P(X=0)=P(X=1)$
<br/><br/>
$$
\begin{aligned}
&3 \cdot{ }^{n} C_{0} P^{0}(1-P)^{n}={ }^{n} C_{1} P^{1}(1-P)^{n-1} \\\\
&\frac{3}{n}=\frac{P}{1-P} \Rightarrow \frac{1}{11}=\frac{P}{1-P} \\\\
&\Rightarrow 1-P=11 P \\\\
&\Rightarrow P=\frac{1}{12}
\end{aligned}
$$
<br/><br/>
$$
\frac{P(X=15)}{P(X=18)}-\frac{P(X=16)}{P(X... | mcq | jee-main-2022-online-24th-june-morning-shift | 7,572 |
1l5w0d935 | maths | probability | binomial-distribution | <p>If a random variable X follows the Binomial distribution B(5, p) such that P(X = 0) = P(X = 1), then $${{P(X = 2)} \over {P(X = 3)}}$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>Given,</p>
<p>$$n = 5$$</p>
<p>and $$P(X = 0) = P(X = 1)$$</p>
<p>We know, $$p(x = r) = {}^n{C_r}\,.\,{p^r}\,.\,{q^{n - r}}$$</p>
<p>where $$p + q = 1$$</p>
<p>$$\therefore$$ $$P(X = 0) = P(X = 1)$$</p>
<p>$$ \Rightarrow {}^5{C_0}\,.\,{p^0}\,.\,{q^5} = {}^5{C_1}\,.\,{p^1}\,.\,{q^4}$$</p>
<p>$$ \Rightarrow 1\,.\,1\,.... | mcq | jee-main-2022-online-30th-june-morning-shift | 7,573 |
1l6dwkw92 | maths | probability | binomial-distribution | <p>If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :
</p> | [{"identifier": "A", "content": "$$\n\\frac{33}{2^{32}}\n$$"}, {"identifier": "B", "content": "$$\\frac{33}{2^{29}}$$"}, {"identifier": "C", "content": "$$\\frac{33}{2^{28}}$$"}, {"identifier": "D", "content": "$$\\frac{33}{2^{27}}$$"}] | ["C"] | null | If $n$ is number of trails, $p$ is probability of success and $q$ is probability of unsuccess then,
<br/><br/>
$$
\begin{aligned}
& \text { Mean }=n p \text { and variance }=n p q \text {. } \\\\
& \text { Here }\\\\
& n p+n p q=24 \quad \dots(i)\\\\
& n p . n p q=128 \quad \dots(ii)\\\\
&\text { and } q=1-p \quad \d... | mcq | jee-main-2022-online-25th-july-morning-shift | 7,574 |
1l6gisnao | maths | probability | binomial-distribution | <p>The mean and variance of a binomial distribution are $$\alpha$$ and $$\frac{\alpha}{3}$$ respectively. If $$\mathrm{P}(X=1)=\frac{4}{243}$$, then $$\mathrm{P}(X=4$$ or 5$$)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{5}{9}$$"}, {"identifier": "B", "content": "$$\\frac{64}{81}$$"}, {"identifier": "C", "content": "$$\\frac{16}{27}$$"}, {"identifier": "D", "content": "$$\\frac{145}{243}$$"}] | ["C"] | null | <p>Given, mean $$ = np = \alpha $$.</p>
<p>and variance $$ = npq = {\alpha \over 3}$$</p>
<p>$$ \Rightarrow q = {1 \over 3}$$ and $$p = {2 \over 3}$$</p>
<p>$$P(X = 1) = n.{p^1}.{q^{n - 1}} = {4 \over {243}}$$</p>
<p>$$ \Rightarrow n.{2 \over 3}.{\left( {{1 \over 3}} \right)^{n - 1}} = {4 \over {243}}$$</p>
<p>$$ \Rig... | mcq | jee-main-2022-online-26th-july-morning-shift | 7,575 |
1l6hz9u96 | maths | probability | binomial-distribution | <p>Let $$X$$ be a binomially distributed random variable with mean 4 and variance $$\frac{4}{3}$$. Then, $$54 \,P(X \leq 2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{73}{27}$$"}, {"identifier": "B", "content": "$$\\frac{146}{27}$$"}, {"identifier": "C", "content": "$$\\frac{146}{81}$$"}, {"identifier": "D", "content": "$$\\frac{126}{81}$$"}] | ["B"] | null | <p>Mean $$ = 4 = \mu = np$$</p>
<p>Variance $$ = {\sigma ^2} = np(1 - P) = {4 \over 3}$$</p>
<p>$$4(1 - P) = {4 \over 3}$$</p>
<p>$$P = {2 \over 3}$$</p>
<p>$$n \times {2 \over 3} = 4$$</p>
<p>$$n = 6$$</p>
<p>$$P(X = k) = {}^n{C_k}\,{P^k}{(1 - P)^{n - k}}$$</p>
<p>$$P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$$</p>
<... | mcq | jee-main-2022-online-26th-july-evening-shift | 7,576 |
1l6kkqt6p | maths | probability | binomial-distribution | <p>Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $$P(X>n-3)=\frac{k}{2^{n}}$$, then k is equal to :</p> | [{"identifier": "A", "content": "528"}, {"identifier": "B", "content": "529"}, {"identifier": "C", "content": "629"}, {"identifier": "D", "content": "630"}] | ["B"] | null | <p>Mean $$ = np = 16$$</p>
<p>Variance $$ = npq = 8$$</p>
<p>$$ \Rightarrow q = p = {1 \over 2}$$ and $$n = 32$$</p>
<p>$$P(x > n - 3) = p(x = n - 2) + p(x = n - 1) + p(x = n)$$</p>
<p>$$ = \left( {{}^{32}{C_2} + {}^{32}{C_1} + {}^{32}{C_0}} \right)\,.\,{1 \over {{2^n}}}$$</p>
<p>$$ = {{529} \over {{2^n}}}$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift | 7,577 |
1l6rffxn0 | maths | probability | binomial-distribution | <p>The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is ____________.</p> | [] | null | 96 | Let two roots of a quadratic equation are mean = np and variance = npq.
<br/><br/>Given $n p+n p q=82.5$
<br/><br/>and $n p(n p q)=1350$
<br/><br/>$$ \therefore $$ Quadratic equation is
<br/><br/>$ x^{2}-82.5 x+1350=0$
<br/><br/>$\Rightarrow x^{2}-22.5 x-60 x+1350=0$
<br/><br/>$\Rightarrow x-(x-22.5)-60(x-22.5)=0$
... | integer | jee-main-2022-online-29th-july-evening-shift | 7,578 |
1ldomop5k | maths | probability | binomial-distribution | <p>In a binomial distribution $$B(n,p)$$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $$6(n+p-q)$$ is equal to :</p> | [{"identifier": "A", "content": "52"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "53"}] | ["A"] | null | $$
\begin{aligned}
& \text { Given } \\\\
& \mathrm{np}+\mathrm{npq}=5 \\\\
& \Rightarrow \mathrm{np}(1+\mathrm{q})=5 ........(i) \\\\
& \text { and (np) (npq) }=6 \\\\
& \Rightarrow \mathrm{n}^2 \mathrm{p}^2 \mathrm{q}=6 ........(ii) \\\\
& (\mathrm{i})^2 \div(\mathrm{ii}) \\\\
& \frac{(1+q)^2}{9}=\frac{25}{6} \\\\
& ... | mcq | jee-main-2023-online-1st-february-morning-shift | 7,579 |
1lgoxr14i | maths | probability | binomial-distribution | <p>The random variable $$\mathrm{X}$$ follows binomial distribution $$\mathrm{B}(\mathrm{n}, \mathrm{p})$$, for which the difference of the mean and the variance is 1 . If $$2 \mathrm{P}(\mathrm{X}=2)=3 \mathrm{P}(\mathrm{X}=1)$$, then $$n^{2} \mathrm{P}(\mathrm{X}>1)$$ is equal to :</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "16"}] | ["C"] | null | $$
\begin{aligned}
& n p-n p q=1 \\\\
\Rightarrow & n p(1-q)=1 \\\\
\Rightarrow & n p^2=1
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& 2 P(X=2)=3 P(X=1) \\\\
& 2 \cdot{ }^n C_2 p^2 q{ }^{n-2}=3 \cdot{ }^n C_1 p \cdot q^{n-1} \\\\
\Rightarrow & 2 \cdot \frac{n \cdot(n-1)}{2} \cdot p=3 \cdot n \cdot q \\\\
\Rightarrow... | mcq | jee-main-2023-online-13th-april-evening-shift | 7,580 |
1lgvphj1r | maths | probability | binomial-distribution | <p> Let a die be rolled $$n$$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $$\frac{k}{2^{15}}$$, then $$\mathrm{k}$$ is equal to :</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "90"}] | ["B"] | null | Given that,
<br/><br/>$\mathrm{P}($ odd number seven times $)=\mathrm{P}($ Odd number nine times)
<br/><br/>$$
\begin{aligned}
& \Rightarrow{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{n-7}={ }^n \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{n-9} \\\\
& \Rightarrow{ }^n \mathr... | mcq | jee-main-2023-online-10th-april-evening-shift | 7,581 |
lv2er9nl | maths | probability | binomial-distribution | <p>In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $$\frac{1}{3}$$ and $$\frac{2}{3}$$ respectively. Let $$x$$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $$\mathrm{P}(|x-y| \leq 2)$$ is $$p$$, then $$3... | [] | null | 8288 | <p>$$\begin{aligned}
& x+y=10 \\
& A=x-y \\
& P(|A|<2) \text { is } P \\
& \Rightarrow|A|=2,1,0 \Rightarrow A=0,1,-1,2,-2 \\
& \Rightarrow x=\frac{10+A}{2} \Rightarrow A \in \text { even as } x \in \text { integer } \\
& \Rightarrow A=0,-2,2 \\
& \Rightarrow P(|A| \leq 2)=P(A=0)+P(A=-2)+P(A=2)
\end{aligned}$$</p>
<p>(1... | integer | jee-main-2024-online-4th-april-evening-shift | 7,583 |
gj3cfkAKO4sn1XEq | maths | probability | classical-defininition-of-probability | Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is : | [{"identifier": "A", "content": "$${{2 \\over 5}}$$"}, {"identifier": "B", "content": "$${{4 \\over 5}}$$"}, {"identifier": "C", "content": "$${{3 \\over 5}}$$"}, {"identifier": "D", "content": "$${{1 \\over 5}}$$"}] | ["A"] | null | <p>X : Mr. A selected wining horse</p>
<p>$$\overline X $$ : Mr. A did not select wining horse</p>
<p>Mr. A selected two horses, now probability of not wining the first horse which Mr. A choses = $${4 \over 5}$$</p>
<p>And probability of not wining the second horse also which Mr. A choses = $${3 \over 4}$$ (Here Mr. A ... | mcq | aieee-2003 | 7,584 |
9sHWR6rtY5FcXa4Y | maths | probability | classical-defininition-of-probability | The probability that $$A$$ speaks truth is $${4 \over 5},$$ while the probability for $$B$$ is $${3 \over 4}.$$ The probability that they contradict each other when asked to speak on a fact is : | [{"identifier": "A", "content": "$${4 \\over 5}$$ "}, {"identifier": "B", "content": "$${1 \\over 5}$$"}, {"identifier": "C", "content": "$${7 \\over 20}$$"}, {"identifier": "D", "content": "$${3 \\over 20}$$"}] | ["C"] | null | <p>The probability of speaking truth by A, P(A) = $${4 \over 5}$$. The probability of not speaking truth by A, P($$\overline A $$) = 1 $$-$$ $${4 \over 5} = {1 \over 5}$$.</p>
<p>The probability of speaking truth by B, P(B) = $${3 \over 4}$$. The probability of not speaking truth of B, P($$\overline B $$) $$ = {1 \over... | mcq | aieee-2004 | 7,585 |
4KumFwAAvt6eWfbO | maths | probability | classical-defininition-of-probability | Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is : | [{"identifier": "A", "content": "$${2 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 9}$$"}, {"identifier": "C", "content": "$${8 \\over 9}$$"}, {"identifier": "D", "content": "$${7 \\over 9}$$"}] | ["B"] | null | <p>Person 1st has three options to apply.</p>
<p>Similarly, person 2nd has three options t apply</p>
<p>and person 3rd has three options to apply.</p>
<p>Total cases = 3<sup>3</sup></p>
<p>Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3)</p>
<p>So, probability $$ = {3 \over {{3^3}}} = {1 \ove... | mcq | aieee-2005 | 7,586 |
K13pUy0l2sZ1pymP | maths | probability | classical-defininition-of-probability | If two different numbers are taken from the set {0, 1, 2, 3, ........, 10}; then the probability that their sum as
well as absolute difference are both multiple of 4, is : | [{"identifier": "A", "content": "$${{12} \\over {55}}$$"}, {"identifier": "B", "content": "$${{14} \\over {45}}$$"}, {"identifier": "C", "content": "$${{7} \\over {55}}$$"}, {"identifier": "D", "content": "$${{6} \\over {55}}$$"}] | ["D"] | null | Let A = {0, 1, 2, 3, 4, ......., 10}
<br><br>Total number of ways of selecting 2 different numbers from A is
<br><br>n (S) = <sup>11</sup>C<sub>2</sub> = 55, where 'S' denotes sample space
<br><br>Let E be the given event
<br><br>$$ \therefore $$ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
<br><br>$$ \Righta... | mcq | jee-main-2017-offline | 7,588 |
e5HmDrCrJDgX09wpKD5Po | maths | probability | classical-defininition-of-probability | In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to : | [{"identifier": "A", "content": "$${{200} \\over {{6^5}}}$$"}, {"identifier": "B", "content": "$${{225} \\over {{6^5}}}$$"}, {"identifier": "C", "content": "$${{150} \\over {{6^5}}}$$"}, {"identifier": "D", "content": "$${{175} \\over {{6^5}}}$$"}] | ["D"] | null | $$\underline {} \,\,\,\underline {} \,\,\,\underline {} \,\,\underline 4 \,\,\underline 4 $$
<br><br>$${1 \over {{6^2}}}\left( {{{{5^3}} \over {{6^3}}} + {{2{C_1}{{.5}^2}} \over {{6^3}}}} \right) = {{175} \over {{6^5}}}$$ | mcq | jee-main-2019-online-12th-january-morning-slot | 7,591 |
m2nwnLqVyIMJTL11MZrgl | maths | probability | classical-defininition-of-probability | The minimum number of times one has to toss a
fair coin so that the probability of observing at least
one head is at least 90% is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["C"] | null | Probablity of getting head P(H) = $${1 \over 2}$$
<br><br>Probablity of getting tail P(T) = 1 - $${1 \over 2}$$ = $${1 \over 2}$$
<br><br>Probability of observing at least one head out of n tosses
<br><br>= 1 - Probability of observing no head occurs out of n tosses
<br><br>= 1 - $${\left( {{1 \over 2}} \right)^n}$$
<b... | mcq | jee-main-2019-online-8th-april-evening-slot | 7,593 |
9n3LPHYPclxu3TKzKw3rsa0w2w9jx22oarv | maths | probability | classical-defininition-of-probability | Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is
more than 99% is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "7"}] | ["D"] | null | $$1 - {\left( {{1 \over 2}} \right)^n} > {{99} \over {100}}$$<br><br>
$${\left( {{1 \over 2}} \right)^n} < {1 \over {100}}$$<br><br>
$$ \Rightarrow $$ n = 7 | mcq | jee-main-2019-online-10th-april-evening-slot | 7,594 |
TlVWYzrt0ahn22L9h7jgy2xukf901fr4 | maths | probability | classical-defininition-of-probability | The probability of a man hitting a target is $${1 \over {10}}$$. The least number of shots required, so that the
probability of his hitting the target at least once is greater than $${1 \over {4}}$$, is ____________. | [] | null | 3 | We have, $$1 - $$(probability of all shots results in failure out of n trials) > $${1 \over 4}$$<br><br>$$ \Rightarrow 1 - {\left( {{9 \over {10}}} \right)^n} > {1 \over 4}$$<br><br>$$ \Rightarrow {3 \over 4} > {\left( {{9 \over {10}}} \right)^n} \Rightarrow n \ge 3$$ | integer | jee-main-2020-online-4th-september-morning-slot | 7,595 |
Hlrmx8ls5tFwjNNkM61klrk41dk | maths | probability | classical-defininition-of-probability | Let B<sub>i</sub> (i = 1, 2, 3) be three independent events in a sample space. The probability that only B<sub>1</sub> occur is $$\alpha $$, only B<sub>2</sub> occurs is $$\beta $$ and only B<sub>3</sub> occurs is $$\gamma $$. Let p be the probability that none of the events B<sub>i</sub> occurs and these 4 probabiliti... | [] | null | 6 | Let x, y, z be probability of B<sub>1</sub>, B<sub>2</sub>, B<sub>3</sub> respectively
<br><br>$$\alpha $$ = P(B<sub>1</sub> $$ \cap $$ $$\overline {{B_2}} \cap \overline {{B_3}} $$) = $$P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$$
<br><br>$$ \Rightarrow $$ x(1 $$-$... | integer | jee-main-2021-online-24th-february-morning-slot | 7,597 |
UBXmKU6xOGiAcbU8No1kmiyynw3 | maths | probability | classical-defininition-of-probability | Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to : | [{"identifier": "A", "content": "$${4 \\over {9}}$$"}, {"identifier": "B", "content": "$${9 \\over {56}}$$"}, {"identifier": "C", "content": "$${11 \\over {27}}$$"}, {"identifier": "D", "content": "$${3 \\over {7}}$$"}] | ["A"] | null | Total cases :<br><br>$$\underline 6 $$ . $$\underline 6 $$ . $$\underline 5 $$ . $$\underline 4 $$ . $$\underline 3 $$ . $$\underline 2 $$<br><br>n(s) = 6 . 6!<br><br>Favourable cases :<br><br>Number divisible by 3 $$ \equiv $$ Sum of digits must be divisible by 3<br><br>Case - I<br><br>1, 2, 3, 4, 5, 6<br><br>Number o... | mcq | jee-main-2021-online-16th-march-evening-shift | 7,599 |
sgzSQk4kJbYhlNtQ1e1kmjazetr | maths | probability | classical-defininition-of-probability | Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is : | [{"identifier": "A", "content": "$${4 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${5 \\over {12}}$$"}, {"identifier": "D", "content": "$${17 \\over {36}}$$"}] | ["D"] | null | n(S) = 36<br><br>possible ordered pair :
<br><br> (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), <br>(2, 1), (2, 2), (2, 3), (2, 5), <br>(3, 1), (3, 2), (3, 3), (3, 5), <br>(5, 1), (5, 2), (5, 3), <br>(7, 1)<br><br>Number of favourable outcomes = 17<br><br>Probability = $${{17} \over {36}}$$ | mcq | jee-main-2021-online-17th-march-morning-shift | 7,600 |
2X5uWEUa0odmBJe9ZR1kmkl6i35 | maths | probability | classical-defininition-of-probability | Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be $${1 \over 2}$$ and probability of occurrence of 0 at the odd place be $${1 \over 3}$$. Then the probability that '10' is followed by '01' is equal to : | [{"identifier": "A", "content": "$${1 \\over 18}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over 9}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}] | ["C"] | null | P(0 at even place) = $${1 \over 2}$$
<br><br>P(0 at odd place) = $${1 \over 3}$$
<br><br> P(1 at even place) = $${1 \over 2}$$
<br><br>P(1 at odd place) = $${2 \over 3}$$
<br><br>P(10 is followed by 01)
<br><br>= $$\left( {{2 \over 3} \times {1 \over 2} \times {1 \over 3} \times {1 \over 2}} \right) + \left( {{1 \over ... | mcq | jee-main-2021-online-17th-march-evening-shift | 7,602 |
1krzrj7ab | maths | probability | classical-defininition-of-probability | A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is ______________. | [] | null | 4 | P(Head) = $${1 \over 2}$$<br><br>1 $$-$$ P(All tail) $$\ge$$ 0.9<br><br>$$1 - {\left( {{1 \over 2}} \right)^n}$$ $$\ge$$ 0.9<br><br>$$ \Rightarrow {\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$$<br><br>$$\Rightarrow$$ n<sub>min</sub> = 4 | integer | jee-main-2021-online-25th-july-evening-shift | 7,604 |
1ks0ai6ce | maths | probability | classical-defininition-of-probability | The probability that a randomly selected 2-digit number belongs to the set {n $$\in$$ N : (2<sup>n</sup> $$-$$ 2) is a multiple of 3} is equal to : | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["C"] | null | Total number of cases = $${}^{90}{C_1} = 90$$<br><br>Now, $${2^n} - 2 = {(3 - 1)^n} - 2$$<br><br>$${}^n{C_0}{3^n} - {}^n{C_1}{.3^{n - 1}} + .... + {( - 1)^{n - 1}}.{}^n{C_{n - 1}}3 + {( - 1)^n}.{}^n{C_n} - 2$$<br><br>= $$3\left( {{3^{n - 1}} - n{3^{n - 2}} + ... + {{( - 1)}^{n - 1}}.n} \right) + {( - 1)^n} - 2$$<br><br... | mcq | jee-main-2021-online-27th-july-morning-shift | 7,605 |
1ktd0dr87 | maths | probability | classical-defininition-of-probability | Two fair dice are thrown. The numbers on them are taken as $$\lambda$$ and $$\mu$$, and a system of linear equations<br/><br/>x + y + z = 5<br/><br/>x + 2y + 3z = $$\mu$$ <br/><br/>x + 3y + $$\lambda$$z = 1<br/><br/>is constructed. If p is the probability that the system has a unique solution and q is the probability t... | [{"identifier": "A", "content": "$$p = {1 \\over 6}$$ and $$q = {1 \\over 36}$$"}, {"identifier": "B", "content": "$$p = {5 \\over 6}$$ and $$q = {5 \\over 36}$$"}, {"identifier": "C", "content": "$$p = {5 \\over 6}$$ and $$q = {1 \\over 36}$$"}, {"identifier": "D", "content": "$$p = {1 \\over 6}$$ and $$q = {5 \\over ... | ["B"] | null | $$D \ne 0 \Rightarrow \left| {\matrix{
1 & 1 & 1 \cr
1 & 2 & 3 \cr
1 & 3 & \lambda \cr
} } \right| \ne 0 \Rightarrow \lambda \ne 5$$<br><br>For no solution D = 0 $$\Rightarrow$$ $$\lambda$$ = 5<br><br>$${D_1} = \left| {\matrix{
1 & 1 & 5 \cr
1 & 2 & \mu... | mcq | jee-main-2021-online-26th-august-evening-shift | 7,606 |
1l56744uu | maths | probability | classical-defininition-of-probability | <p>The probability, that in a randomly selected 3-digit number at least two digits are odd, is :</p> | [{"identifier": "A", "content": "$${{19} \\over {36}}$$"}, {"identifier": "B", "content": "$${{15} \\over {36}}$$"}, {"identifier": "C", "content": "$${{13} \\over {36}}$$"}, {"identifier": "D", "content": "$${{23} \\over {36}}$$"}] | ["A"] | null | At least two digits are odd = exactly two digits are odd + exactly there 3 digits
are odd <br><br>For exactly three digits are odd <br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8eefiq/e8de7882-e991-47ff-999d-2ea9ce2e3c3f/ccd33f20-8716-11ed-92bb-e57e64e1a06d/file-1lc8eefir.png?format=png" d... | mcq | jee-main-2022-online-28th-june-morning-shift | 7,607 |
1l56rh8nv | maths | probability | classical-defininition-of-probability | <p>If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x $$-$$ 6y = 30, then the probability that y < 1 is :</p> | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${5 \\over 6}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${6 \\over 7}$$"}] | ["B"] | null | <p>The required probability</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7g5yy/9381a58a-1f21-4b26-abca-dc94a3e72b11/6c8ce9a0-05bf-11ed-8617-d71e6444d1a0/file-1l5p7g5yz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5p7g5yy/9381a58a-1f21-4b26-abca-dc94a3e72b11/6c8c... | mcq | jee-main-2022-online-27th-june-evening-shift | 7,608 |
1l56uaobr | maths | probability | classical-defininition-of-probability | <p>Let S = {E<sub>1</sub>, E<sub>2</sub>, ........., E<sub>8</sub>} be a sample space of a random experiment such that $$P({E_n}) = {n \over {36}}$$ for every n = 1, 2, ........, 8. Then the number of elements in the set $$\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$$ is ___________.</p> | [] | null | 19 | <p>Here $$P({E_n}) = {n \over {36}}$$ for n = 1, 2, 3, ......, 8</p>
<p>Here $$P(A) = {{Any\,possible\,sum\,of\,(1,2,3,\,...,\,8)( = a\,say)} \over {36}}$$</p>
<p>$$\because$$ $${a \over {36}} \ge {4 \over 5}$$</p>
<p>$$\therefore$$ $$a \ge 29$$</p>
<p>If one of the number from {1, 2, ......, 8} is left then total $$a ... | integer | jee-main-2022-online-27th-june-evening-shift | 7,609 |
1l6dwn5ao | maths | probability | classical-defininition-of-probability | <p>If the numbers appeared on the two throws of a fair six faced die are $$\alpha$$ and $$\beta$$, then the probability that $$x^{2}+\alpha x+\beta>0$$, for all $$x \in \mathbf{R}$$, is :
</p> | [{"identifier": "A", "content": "$$\\frac{17}{36}$$"}, {"identifier": "B", "content": "$$\n\\frac{4}{9}\n$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$\\frac{19}{36}$$"}] | ["A"] | null | For $x^{2}+\alpha x+\beta>0 \forall x \in R$ to hold, we should have $\alpha^{2}-4 \beta<0$
<br/><br/>
If $\alpha=1, \beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices
<br/><br/>
If $\alpha=2, \beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices
<br/><br/>
If $\alpha=3, \beta$ can be $3,4,5,6$ i.e., 4 choices
<br/><br/>
If $\alpha=... | mcq | jee-main-2022-online-25th-july-morning-shift | 7,610 |
1l6jc7ci7 | maths | probability | classical-defininition-of-probability | <p>Let $$S$$ be the sample space of all five digit numbers. It $$p$$ is the probability that a randomly selected number from $$S$$, is a multiple of 7 but not divisible by 5 , then $$9 p$$ is equal to :</p> | [{"identifier": "A", "content": "1.0146"}, {"identifier": "B", "content": "1.2085"}, {"identifier": "C", "content": "1.0285"}, {"identifier": "D", "content": "1.1521"}] | ["C"] | null | <p>Among the 5 digit numbers,</p>
<p>First number divisible by 7 is 10003 and last is 99995.</p>
<p>$$\Rightarrow$$ Number of numbers divisible by 7.</p>
<p>$$ = {{99995 - 10003} \over 7} + 1$$</p>
<p>$$ = 12857$$</p>
<p>First number divisible by 35 is 10010 and last is 99995.</p>
<p>$$\Rightarrow$$ Number of numbers d... | mcq | jee-main-2022-online-27th-july-morning-shift | 7,611 |
ldoatpci | maths | probability | classical-defininition-of-probability | Let A be the event that the absolute difference between two randomly choosen real numbers in the sample space $[0,60]$ is less than or equal to a . If $\mathrm{P}(\mathrm{A})=\frac{11}{36}$, then $\mathrm{a}$ is equal to _______. | [] | null | 10 | $$
\begin{aligned}
& |\mathrm{x}-\mathrm{y}|<\mathrm{a} \Rightarrow-\mathrm{a}<\mathrm{x}-\mathrm{y}<\mathrm{a} \\\\
& \Rightarrow \mathrm{x}-\mathrm{y}<\mathrm{a} \text { and } \mathrm{x}-\mathrm{y}>-\mathrm{a}
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/im... | integer | jee-main-2023-online-31st-january-evening-shift | 7,612 |
ldr0ncc0 | maths | probability | classical-defininition-of-probability | A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is p. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is $q$. If $p: q=m: n$, where $... | [] | null | 14 | <p>$$p = {6 \over {36}} = {1 \over 6}$$</p>
<p>$$q = {{{}^6{C_1} \times {}^5{C_1} \times {{4!} \over {3!}}} \over {{6^4}}} = {{120} \over {1296}} = {5 \over {54}}$$</p>
<p>$${p \over q} = {{{1 \over 6}} \over {{5 \over {54}}}} = {{54} \over {6 \times 5}} = {9 \over 5} = {m \over n}$$</p>
<p>$$m + n = 14$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 7,613 |
1ldv1kqtg | maths | probability | classical-defininition-of-probability | <p>Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $$S = \left\{ {x \in \mathbb{Z}:x(66 - x) \ge {5 \over 9}M} \right\}$$ and the event $$\mathrm{A = \{ x \in S:x\,is\,a\,multiple\,of\,3\}}$$. Then P(A) is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{3}$$"}, {"identifier": "B", "content": "$$\\frac{1}{5}$$"}, {"identifier": "C", "content": "$$\\frac{7}{22}$$"}, {"identifier": "D", "content": "$$\\frac{15}{44}$$"}] | ["A"] | null | $x+y=66$
<br/><br/>
$$
\begin{aligned}
& \frac{x+y}{2} \geq \sqrt{x y} \\\\
\Rightarrow & 33 \geq \sqrt{x y} \\\\
\Rightarrow & x y \leq 1089 \\\\
\therefore & M=1089 \\\\
S: & x(66-x) \geq \frac{5}{9} \cdot 1089 \\\\
& 66 x-x^{2} \geq 605 \\\\
\Rightarrow & x^{2}-66 x+605 \leq 0
\end{aligned}
$$
<br/><br/>$\Rightarrow... | mcq | jee-main-2023-online-25th-january-morning-shift | 7,616 |
1ldyaqkkl | maths | probability | classical-defininition-of-probability | <p>Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations</p>
<p>$$x + y + z = 1$$</p>
<p>$$2x + \mathrm{N}y + 2z = 2$$</p>
<p>$$3x + 3y + \mathrm{N}z = 3$$</p>
<p>has unique solution is $${k \over 6}$$, then the sum of value of k and all possible values of N is... | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "19"}] | ["C"] | null | For unique solution $\Delta \neq 0$
<br/><br/>
i.e. $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{array}\right| \neq 0$
<br/><br/>
$\Rightarrow\left(N^{2}-6\right)-(2 N-6)+(6-3 N) \neq 0$
<br/><br/>
$\Rightarrow N^{2}-5 N+6 \neq 0$
<br/><br/>
$\therefore N \neq 2$ and $N \neq 3$
<br/><br/>
$\therefor... | mcq | jee-main-2023-online-24th-january-morning-shift | 7,617 |
1ldyb4hyp | maths | probability | classical-defininition-of-probability | <p>Let $$\Omega$$ be the sample space and $$\mathrm{A \subseteq \Omega}$$ be an event.</p>
<p>Given below are two statements :</p>
<p>(S1) : If P(A) = 0, then A = $$\phi$$</p>
<p>(S2) : If P(A) = 1, then A = $$\Omega$$</p>
<p>Then :</p> | [{"identifier": "A", "content": "both (S1) and (S2) are true"}, {"identifier": "B", "content": "both (S1) and (S2) are false"}, {"identifier": "C", "content": "only (S2) is true"}, {"identifier": "D", "content": "only (S1) is true"}] | ["B"] | null | $\Omega=$ sample space<br/><br/>
$\mathrm{A}=$ be an event<br/><br/>
$ \Omega$ = A wire of length 1 which starts at point 0 and ends at point 1 on the coordinate axis = $[0,1]$<br/><br/>
$\mathrm{A}=\left\{\frac{1}{2}\right\}$ = Selecting a point on the wire which is at $\left\{\frac{1} {2}\right\}$ or 0.5
<br/><br/>A... | mcq | jee-main-2023-online-24th-january-morning-shift | 7,618 |
1lgutw8tu | maths | probability | classical-defininition-of-probability | <p>Let $$S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$$ be a sample space and $$A=\{M \in S: M$$ is invertible $$\}$$ be an event. Then $$P(A)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{47}{81}$$"}, {"identifier": "B", "content": "$$\\frac{49}{81}$$"}, {"identifier": "C", "content": "$$\\frac{50}{81}$$"}, {"identifier": "D", "content": "$$\\frac{16}{27}$$"}] | ["C"] | null | We have, $S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$
<br/><br/>Let $M=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, where $a, b, c, d \in\{0,1,2\}$
<br/><br/>$$
n(s)=3^4=81
$$
<br/><br/>If $A$ is invertible, then $|A| \neq 0$
<br/><br/>Now, if $|A|=0$, then $|M|=0$
<b... | mcq | jee-main-2023-online-11th-april-morning-shift | 7,619 |
1lgxt6jpu | maths | probability | classical-defininition-of-probability | <p>Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that $${2^N} < N!$$ is $${m \over n}$$, where m and n are coprime, then $$4m-3n$$ is equal to :</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "10"}] | ["C"] | null | $N$ denote the sum of the numbers obtained when two dice are rolled.
<br/><br/>Such that $2^N < N$!
<br/><br/>$$
\text { i.e., } 4 \leq N \leq 12 \text { i.e., } N \in\{4,5,6, \ldots 12\}
$$
<br/><br/>Now, $P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$
<br/><br/>So, required probability $=1-\frac{1... | mcq | jee-main-2023-online-10th-april-morning-shift | 7,620 |
jaoe38c1lsfkotb7 | maths | probability | classical-defininition-of-probability | <p>An integer is chosen at random from the integers $$1,2,3, \ldots, 50$$. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is</p> | [{"identifier": "A", "content": "$$\\frac{8}{25}$$\n"}, {"identifier": "B", "content": "$$\\frac{9}{50}$$\n"}, {"identifier": "C", "content": "$$\\frac{14}{25}$$\n"}, {"identifier": "D", "content": "$$\\frac{21}{50}$$"}] | ["D"] | null | <p>Given set $$=\{1,2,3, \ldots \ldots . .50\}$$</p>
<p>$$\mathrm{P}(\mathrm{A})=$$ Probability that number is multiple of 4</p>
<p>$$\mathrm{P(B)}=$$ Probability that number is multiple of 6</p>
<p>$$\mathrm{P}(\mathrm{C})=$$ Probability that number is multiple of 7</p>
<p>Now,</p>
<p>$$\mathrm{P}(\mathrm{A})=\frac{12... | mcq | jee-main-2024-online-29th-january-evening-shift | 7,622 |
1lsgadxd6 | maths | probability | classical-defininition-of-probability | <p>Two integers $$x$$ and $$y$$ are chosen with replacement from the set $$\{0,1,2,3, \ldots, 10\}$$. Then the probability that $$|x-y|>5$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{31}{121}$$\n"}, {"identifier": "B", "content": "$$\\frac{60}{121}$$\n"}, {"identifier": "C", "content": "$$\\frac{62}{121}$$\n"}, {"identifier": "D", "content": "$$\\frac{30}{121}$$"}] | ["D"] | null | <p>If $$x=0, y=6,7,8,9,10$$</p>
<p>If $$x=1, y=7,8,9,10$$</p>
<p>If $$x=2, y=8,9,10$$</p>
<p>If $$x=3, y=9,10$$</p>
<p>If $$x=4, y=10$$</p>
<p>If $$x=5, y=$$ no possible value</p>
<p>Total possible ways $$=(5+4+3+2+1) \times 2$$</p>
<p>$$=30$$</p>
<p>Required probability $$=\frac{30}{11 \times 11}=\frac{30}{121}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 7,623 |
luxwcbol | maths | probability | classical-defininition-of-probability | <p>If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $$i^{\text {th }}$$ roll than the number obtained in the $$(i-1)^{\text {th }}$$ roll, $$i=2,3$$, is equal to</p> | [{"identifier": "A", "content": "5/54"}, {"identifier": "B", "content": "2/54"}, {"identifier": "C", "content": "1/54"}, {"identifier": "D", "content": "3/54"}] | ["A"] | null | <p>Let's denote the outcomes of the three rolls as $$X_1$$, $$X_2$$, and $$X_3$$, where $$X_i$$ represents the number obtained in the $$i^{\text{th}}$$ roll. We are looking for the probability that:</p>
<p>$$X_2 > X_1 \text{ and } X_3 > X_2$$</p>
<p>The total number of outcomes when rolling a dice three times is: </p... | mcq | jee-main-2024-online-9th-april-evening-shift | 7,624 |
luy9clq6 | maths | probability | classical-defininition-of-probability | <p>Let $$\mathrm{a}, \mathrm{b}$$ and $$\mathrm{c}$$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $$1,2,3,4$$. If the probability that $$a x^2+b x+c=0$$ has all real roots is $$\frac{m}{n}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$, then $$\mathrm{m}+\mathrm{... | [] | null | 19 | <p>A quadratic equation $$ax^2 + bx + c = 0$$ has real roots if and only if its discriminant is non-negative. The discriminant $$\Delta$$ of the quadratic equation is given by:</p>
<p>$$\Delta = b^2 - 4ac$$</p>
<p>For the quadratic equation to have all real roots, the discriminant must be non-negative:</p>
<p>$$\Del... | integer | jee-main-2024-online-9th-april-morning-shift | 7,625 |
lv5grw62 | maths | probability | classical-defininition-of-probability | <p>Let the sum of two positive integers be 24 . If the probability, that their product is not less than $$\frac{3}{4}$$ times their greatest possible product, is $$\frac{m}{n}$$, where $$\operatorname{gcd}(m, n)=1$$, then $$n$$-$$m$$ equals</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>Take two numbers as $$a$$ and $$b$$</p>
<p>$$a+b=24$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8pero8/601e15c3-bb4a-487e-b3ff-b7b116b4f031/2e363180-1336-11ef-9f8d-838c388c326d/file-1lw8pero9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8pero8/601e15c3-bb4... | mcq | jee-main-2024-online-8th-april-morning-shift | 7,626 |
lv7v4fz3 | maths | probability | classical-defininition-of-probability | <p>The coefficients $$a, b, c$$ in the quadratic equation $$a x^2+b x+c=0$$ are chosen from the set $$\{1,2,3,4,5,6,7,8\}$$. The probability of this equation having repeated roots is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{128}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{64}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{256}$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{128}$$"}] | ["B"] | null | <p>Given quadratic equation is</p>
<p>$$a x^2+b x+c=0 \text { where } a, b, c \in\{1,2,3, \ldots, 8\}$$</p>
<p>For repeated roots,</p>
<p>$$\begin{aligned}
& b^2-4 a c=0 \\
& \Rightarrow b^2=4 a c
\end{aligned}$$</p>
<p>$$\Rightarrow a c$$ must be a perfect square</p>
<p>$$(a, c) \in\{(1,1),(1,4),(2,2),(2,8),(3,3),(4,1... | mcq | jee-main-2024-online-5th-april-morning-shift | 7,627 |
lvb294g8 | maths | probability | classical-defininition-of-probability | <p>If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is :</p> | [{"identifier": "A", "content": "$$\\frac{18}{25}$$\n"}, {"identifier": "B", "content": "$$\\frac{12}{25}$$\n"}, {"identifier": "C", "content": "$$\\frac{6}{25}$$\n"}, {"identifier": "D", "content": "$$\\frac{4}{25}$$"}] | ["B"] | null | <p>We have 3 letters and 5 addresses, where 3 letters are posted to exactly 2 addresses. First, we will select 2 addresses in $${ }^5 C_2$$ ways.</p>
<p>Now, 3 letters can be posted to exactly 2 addresses in 6 ways.</p>
<p>$$\begin{aligned}
\therefore \text { Probability }=\frac{{ }^5 C_2 \times 6}{5^3} & \\
& =\frac{6... | mcq | jee-main-2024-online-6th-april-evening-shift | 7,628 |
cR0wVgMAi5qt4x9Q | maths | probability | conditional-probability-and-multiplication-theorem | Two aeroplanes $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ bomb a target in succession. The probabilities of $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ scoring a hit correctly are $$0.3$$ and $$0.2,$$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the seco... | [{"identifier": "A", "content": "$$0.2$$ "}, {"identifier": "B", "content": "$$0.7$$ "}, {"identifier": "C", "content": "$$0.06$$ "}, {"identifier": "D", "content": "0.32"}] | ["D"] | null | <p>The desired probability</p>
<p>= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + .......</p>
<p>= 0.14 [1 + (0.56) + (0.56)<sup>2</sup> + .......]</p>
<p>= 0.14 $$\left( {{1 \over {1 - 0.56}}} \right) = {{0.14} \over {0.44}} = {7 \over {22}}$$ = 0.32</p> | mcq | aieee-2007 | 7,629 |
9tucb0kVsi88la1U | maths | probability | conditional-probability-and-multiplication-theorem | It is given that the events $$A$$ and $$B$$ are such that
<br/>$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is : | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | Given that,
<br><br>$$P\left( {{A \over B}} \right) = {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$$ = $${1 \over 2}$$.............. equation (1)
<br><br>$$P\left( {{B \over A}} \right) = {2 \over 3}$$
<br><br>$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over ... | mcq | aieee-2008 | 7,630 |
29tfdggsNkdMWL3J | maths | probability | conditional-probability-and-multiplication-theorem | If $$C$$ and $$D$$ are two events such that $$C \subset D$$ and $$P\left( D \right) \ne 0,$$ then the correct statement among the following is : | [{"identifier": "A", "content": "$$P\\left( {{C \\over D}} \\right)$$$$ \\ge P\\left( C \\right)$$ "}, {"identifier": "B", "content": "$$P\\left( {{C \\over D}} \\right)$$$$ < P\\left( C \\right)$$ "}, {"identifier": "C", "content": "$$P\\left( {{C \\over D}} \\right)$$$$ = {{P\\left( D \\right)} \\over {P\\left( C ... | ["A"] | null | Given that $$C \subset D$$ means $$C$$ is present entirely inside $$D$$. Which is shown below.
<img class="question-image" src="https://imagex.cdn.examgoal.net/Fyl9Vf0KWiyB0u0XJ/tctGFeitNT0FkdGQUNU4VngNUHo8n/XoAkkZjSps5tTrqaC7VVGN/image.svg" loading="lazy" alt="AIEEE 2011 Mathematics - Probability Question 174 English ... | mcq | aieee-2011 | 7,632 |
gTCUP1krIS6fyFBA | maths | probability | conditional-probability-and-multiplication-theorem | Let $$A$$ and $$B$$ be two events such that $$P\left( {\overline {A \cup B} } \right) = {1 \over 6},\,P\left( { {A \cap B} } \right) = {1 \over 4}$$ and $$P\left( {\overline A } \right) = {1 \over 4},$$ where $$\overline A $$ stands for the complement of the event $$A$$. Then the events $$A$$ and $$B$$ are : | [{"identifier": "A", "content": "independent but not equally likely. "}, {"identifier": "B", "content": "independent and equally likely. "}, {"identifier": "C", "content": "mutually exclusive and independent."}, {"identifier": "D", "content": "equally likely but not independent."}] | ["A"] | null | <p>$$P(\overline {A \cup B} ) = {1 \over 6}$$</p>
<p>or, $$1 - P(A \cup B) = {1 \over 6}$$</p>
<p>$$\therefore$$ $$P(A \cup B) = 1 - {1 \over 6} = {5 \over 6};$$ $$P(A \cap B) = {1 \over 4}$$; and $$P(\overline A ) = {1 \over 4};$$</p>
<p>$$\therefore$$ $$P(A) = 1 - P(\overline A ) = 1 - {1 \over 4} = {3 \over 4}$$</p>... | mcq | jee-main-2014-offline | 7,634 |
IEceFTmzpblWyxmi | maths | probability | conditional-probability-and-multiplication-theorem | Let two fair six-faced dice $$A$$ and $$B$$ be thrown simultaneously. If $${E_1}$$ is the event that die $$A$$ shows up four, $${E_2}$$ is the event that die $$B$$ shows up two and $${E_3}$$ is the event that the sum of numbers on both dice is odd, then which of the following statements is $$NOT$$ true? | [{"identifier": "A", "content": "$${E_1}$$ and $${E_2}$$ are independent."}, {"identifier": "B", "content": "$${E_2}$$ and $${E_3}$$ are independent."}, {"identifier": "C", "content": "$${E_1}$$ and $${E_3}$$ are independent."}, {"identifier": "D", "content": "$${E_1},$$ $${E_2}$$ and $${E_3}$$ are independent."}] | ["D"] | null | Total possible outcome with two six faced dice = 6<sup>2</sup> = 36
<br><br>When dice A shows up 4, the possible cases are
<br>E<sub>1</sub> = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
<br>$$\therefore$$ $$P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}$$
<br><br> When dice B shows up 2, the possib... | mcq | jee-main-2016-offline | 7,635 |
PTFEW83CW5cQQ7CvoX4Zl | maths | probability | conditional-probability-and-multiplication-theorem | If A and B are any two events such that P(A) = $${2 \over 5}$$ and P (A $$ \cap $$ B) = $${3 \over {20}}$$, hen the conditional probability, P(A $$\left| {} \right.$$(A' $$ \cup $$ B')), where A' denotes the complement of A, is equal to : | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${5 \\over 17}$$"}, {"identifier": "C", "content": "$${8 \\over 17}$$"}, {"identifier": "D", "content": "$${11 \\over 20}$$"}] | ["B"] | null | $$P\left( {{A \over {A' \cup B'}}} \right)$$
<br><br>$$ = {{P\left[ {A \cap \left( {A' \cup B'} \right)} \right]} \over {P\left( {A' \cap B'} \right)}}$$
<br><br>$$ = {{P\left[ {\left( {A \cap A'} \right) \cup \left( {A \cap B'} \right)} \right]} \over {P\left( {A \cap B} \right)'}}$$
<br><br>$$\left[ \, \right.$$As&nb... | mcq | jee-main-2016-online-9th-april-morning-slot | 7,636 |
GzgdjCK46V8fjwqTu9deK | maths | probability | conditional-probability-and-multiplication-theorem | Let E and F be two independent events. The probability that both E and F happen is $${1 \over {12}}$$ and the probability that neither E nor F happens is $${1 \over {2}}$$, then a value of $${{P\left( E \right)} \over {P\left( F \right)}}$$ is : | [{"identifier": "A", "content": "$${4 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 3}$$"}, {"identifier": "D", "content": "$${5 \\over 12}$$"}] | ["A"] | null | <p>Let P(E) = x and P(F) = y</p>
<p>Now, $$P(E \cap F) = {1 \over {12}}$$</p>
<p>$$ \Rightarrow P(E)P(F) = {1 \over {12}}$$</p>
<p>$$ \Rightarrow xy = {1 \over {12}}$$</p>
<p>Also, $$P(E' \cap F') = {1 \over 2}$$</p>
<p>$$ \Rightarrow (1 - P(E))(1 - P(F)) = {1 \over 2}$$</p>
<p>$$ \Rightarrow (1 - x)(1 - y) = {1 \over ... | mcq | jee-main-2017-online-9th-april-morning-slot | 7,637 |
QpF3mzdVaPsLYadpaz830 | maths | probability | conditional-probability-and-multiplication-theorem | A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the v... | [{"identifier": "A", "content": "$${1 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["B"] | null | P(X getting head) = p
<br><br>$$ \therefore $$ P(X getting tail) = 1 - p
<br><br>P(Y getting head) = P(Y getting tail) = $${1 \over 2}$$
<br><br>P(X wins) = p + (1 - p)$${1 \over 2}$$p + (1 - p)$${1 \over 2}$$(1 - p)$${1 \over 2}$$p + ...
<br><br>= $${p \over {1 - \left( {{{1 - p} \over 2}} \right)}}$$
<br><br>= $${{2p... | mcq | jee-main-2018-online-15th-april-evening-slot | 7,638 |
pKPACm03fi1d3b1OacMOk | maths | probability | conditional-probability-and-multiplication-theorem | Let A, B and C be three events, which are pair-wise independent and $$\overrightarrow E $$ denotes the completement of an event E. If $$P\left( {A \cap B \cap C} \right) = 0$$ and $$P\left( C \right) > 0,$$ then $$P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right]$$ is equal to : | [{"identifier": "A", "content": "$$P\\left( {\\overline A } \\right) - P\\left( B \\right)$$"}, {"identifier": "B", "content": "$$P\\left( A \\right) + P\\left( {\\overline B } \\right)$$"}, {"identifier": "C", "content": "$$P\\left( {\\overline A } \\right) - P\\left( {\\overline B } \\right)$$"}, {"identifier": "D", ... | ["A"] | null | Here, $$P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}$$
<br><br>= $${{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}$$
<br><br>= $${{P\left[ {C - \left( {A \cup B} \rig... | mcq | jee-main-2018-online-16th-april-morning-slot | 7,639 |
YsZBablbi4cftZyZiVjNo | maths | probability | conditional-probability-and-multiplication-theorem | An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probabili... | [{"identifier": "A", "content": "$${{21} \\over {49}}$$"}, {"identifier": "B", "content": "$${{27} \\over {49}}$$"}, {"identifier": "C", "content": "$${{26} \\over {49}}$$"}, {"identifier": "D", "content": "$${{32} \\over {49}}$$"}] | ["D"] | null | 5 Red and 2 green balls
<br><br>P(one red ball) = $${5 \over 7}$$
<br><br>P(one green ball) = $${2 \over 7}$$
<br><br><b>Case I : </b>
<br><br>If drawn ball is green than a red ball is added
<br><br>$$\left( {\matrix{
{6{\mathop{\rm Re}\nolimits} d} \cr
{1\,Green} \cr
} } \right)$$ P (red ball) = $${6 \over... | mcq | jee-main-2019-online-9th-january-evening-slot | 7,640 |
bvonEScgOPynV9McufeFA | maths | probability | conditional-probability-and-multiplication-theorem | An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ……, 9 is randomly picked and the number on the card is noted. Th... | [{"identifier": "A", "content": "$${{19} \\over {36}}$$"}, {"identifier": "B", "content": "$${{15} \\over {72}}$$"}, {"identifier": "C", "content": "$${{13} \\over {36}}$$"}, {"identifier": "D", "content": "$${{19} \\over {72}}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265335/exam_images/oszgg4ddvwq2jkpn9psx.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Probability Question 153 English Explanation">
<br>$$P\l... | mcq | jee-main-2019-online-10th-january-morning-slot | 7,641 |
zrFBvUWrqtdjgdeYy5zha | maths | probability | conditional-probability-and-multiplication-theorem | Let A and B be two non-null events such that
A $$ \subset $$ B . Then, which of the following statements
is always correct? | [{"identifier": "A", "content": "P(A|B) = 1"}, {"identifier": "B", "content": "P(A|B) = P(B) \u2013 P(A)"}, {"identifier": "C", "content": "P(A|B) $$ \\le $$ P(A)\n"}, {"identifier": "D", "content": "P(A|B) $$ \\ge $$ P(A)\n"}] | ["D"] | null | $$P\left( {{A \over B}} \right) = {{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$$
<br><br>As A $$ \subset $$ B,
<br><br>then P(A$$ \cap $$B) = P(A)
<br><br>$$ \therefore $$ $$P\left( {{A \over B}} \right) = {{P\left( A \right)} \over {P\left( B \right)}}$$
<br><br>As P(B) $$ \le $$ 1
<br><br>$$ \therefore $$... | mcq | jee-main-2019-online-8th-april-morning-slot | 7,643 |
FtUibW67JlXKrCLOmU18hoxe66ijvwpbb2e | maths | probability | conditional-probability-and-multiplication-theorem | Four persons can hit a target correctly with
probabilities
$${1 \over 2}$$, $${1 \over 3}$$, $${1 \over 4}$$ and
$${1 \over 8}$$ respectively. if all hit
at the target independently, then the probability that
the target would be hit, is : | [{"identifier": "A", "content": "$${{25} \\over {32}}$$"}, {"identifier": "B", "content": "$${{25} \\over {192}}$$"}, {"identifier": "C", "content": "$${{1} \\over {192}}$$"}, {"identifier": "D", "content": "$${{7} \\over {32}}$$"}] | ["A"] | null | Let four persons are A, B, C and D.
<br><br>Probablity of hitting a target by them,
<br><br>P(A) = $${1 \over 2}$$
<br><br>P(B) = $${1 \over 3}$$
<br><br>P(C) = $${1 \over 4}$$
<br><br>P(D) = $${1 \over 8}$$
<br><br>Probablity of hitting target atleast once = 1 - Probablity of not hitting by anybody
<br><br>P(Hit) = 1 ... | mcq | jee-main-2019-online-9th-april-morning-slot | 7,644 |
0lS1ulEsuyoAXjx4Pr3rsa0w2w9jwxkwc5u | maths | probability | conditional-probability-and-multiplication-theorem | Assume that each born child is equally likely to be a boy or a girl. If two families have two children each,
then the conditional probability that all children are girls given that at least two are girls is : | [{"identifier": "A", "content": "$${1 \\over {10}}$$"}, {"identifier": "B", "content": "$${1 \\over {17}}$$"}, {"identifier": "C", "content": "$${1 \\over {11}}$$"}, {"identifier": "D", "content": "$${1 \\over {12}}$$"}] | ["C"] | null | A = At least two girls<br><br>
B = All girls<br><br>
$$P\left( {{B \over A}} \right) = {{P\left( {B \cap A} \right)} \over {P\left( A \right)}}$$<br><br>
$$ \Rightarrow {{P(B)} \over {P(A)}} = {{{{\left( {{1 \over 4}} \right)}^2}} \over {1 - {}^4{C_0}{{\left( {{1 \over 2}} \right)}^4} - {}^4{C_1}{{\left( {{1 \over 2}}... | mcq | jee-main-2019-online-10th-april-morning-slot | 7,645 |
WIVZ5wahzNzL3rDRF2jgy2xukfajvp4w | maths | probability | conditional-probability-and-multiplication-theorem | In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws total a of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops... | [{"identifier": "A", "content": "$${5 \\over {6}}$$"}, {"identifier": "B", "content": "$${5 \\over {31}}$$"}, {"identifier": "C", "content": "$${31 \\over {61}}$$"}, {"identifier": "D", "content": "$${30 \\over {61}}$$"}] | ["D"] | null | Sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}
<br><br>$$P(6) = {5 \over 36}$$
<br><br>Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}
<br><br>$$\,P(7) = {6 \over {36}}$$ = $${1 \over 6}$$
<br><br>Game ends and A wins if A throws 6 in 1<sup>st</sup>
throw or A don’t throw 6 in 1<sup>st</sup> throw, B don’t
throw 7 in 1<sup>... | mcq | jee-main-2020-online-4th-september-evening-slot | 7,646 |
1YfQyWb3UUUTNEknDmjgy2xukezeh1np | maths | probability | conditional-probability-and-multiplication-theorem | Let E<sup>C</sup> denote the complement of an event E.
Let E<sub>1</sub>
, E<sub>2</sub>
and E<sub>3</sub>
be any pairwise independent
events with P(E<sub>1</sub>) > 0
<br/><br>and P(E<sub>1</sub> $$ \cap $$ E<sub>2</sub> $$ \cap $$ E<sub>3</sub>) = 0.
<br/><br>Then P($$E_2^C \cap E_3^C/{E_1}$$) is equal to :</br... | [{"identifier": "A", "content": "$$P\\left( {E_3^C} \\right)$$ - P(E<sub>2</sub>)"}, {"identifier": "B", "content": "$$P\\left( {E_2^C} \\right)$$ + P(E<sub>3</sub>)"}, {"identifier": "C", "content": "$$P\\left( {E_3^C} \\right)$$ - $$P\\left( {E_2^C} \\right)$$"}, {"identifier": "D", "content": "P(E<sub>3</sub>) - $$P... | ["A"] | null | Given E<sub>1</sub> , E<sub>2</sub> , E<sub>3</sub> are pairwise indepedent events
<br><br>so P(E<sub>1</sub> $$ \cap $$ E<sub>2</sub> ) = P(E<sub>1</sub> ).P(E<sub>2</sub> )
<br><br>and P(E<sub>2</sub> $$ \cap $$ E<sub>3</sub> ) = P(E<sub>2</sub> ).P(E<sub>3</sub> )
<br><br>and P(E<sub>3</sub> $$ \cap $$ E<sub>1</... | mcq | jee-main-2020-online-2nd-september-evening-slot | 7,647 |
bsMRfGHLGyb32Fe3n2jgy2xukezlxh5x | maths | probability | conditional-probability-and-multiplication-theorem | A dice is thrown two times and the sum of the
scores appearing on the die is observed to be
a multiple of 4. Then the conditional probability
that the score 4 has appeared atleast once is : | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "$${1 \\over 9}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | Let A is the event for getting score a multiple
of 4.
<br><br>So, A = { (1, 3), (3, 1), (2, 2), (2, 6), (6, 2),
(3, 5), (5, 3), (4, 4), (6, 6) } = 09
<br><br>n(A) = 9
<br><br>B : Score of 4 has appeared at least once.
<br><br>B = {(4, 4)}
<br><br>So, Required probability = $${1 \over 9}$$ | mcq | jee-main-2020-online-3rd-september-morning-slot | 7,649 |
aXHQRXYYpHINc0geBR1kls3ppy8 | maths | probability | conditional-probability-and-multiplication-theorem | When a missile is fired from a ship, the probability that it is intercepted is $${1 \over 3}$$ and the probability that the missile hits the target, given that it is not intercepted, is $${3 \over 4}$$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${3 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 27}$$"}, {"identifier": "D", "content": "$${1 \\over 8}$$"}] | ["D"] | null | Probability of not getting intercepted = $${2 \over 3}$$<br><br>When it is not intercepted, probability of missile hitting target = $${3 \over 4}$$<br><br>$$\therefore$$ So when such 3 missiles launched
then P (all 3 hitting the target)
<br><br>= $${\left( {{2 \over 3} \times {3 \over 4}} \right)} $$ $$ \times $$ $${\... | mcq | jee-main-2021-online-25th-february-morning-slot | 7,650 |
lNV17llKqZw6mSaHaj1kmhz6wk9 | maths | probability | conditional-probability-and-multiplication-theorem | A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is : | [{"identifier": "A", "content": "$${{39} \\over {50}}$$"}, {"identifier": "B", "content": "$${{3} \\over {4}}$$"}, {"identifier": "C", "content": "$${{22} \\over {425}}$$"}, {"identifier": "D", "content": "$${{52} \\over {867}}$$"}] | ["A"] | null | Consider the events,
<br><br>E<sub>1</sub> = missing card is spade
<br><br>E<sub>2</sub> = missing card is not a spade
<br><br>A = Two spade cards are drawn
<br><br>$$P\left( {{E_1}} \right) = {1 \over 4}$$
<br>$$P\left( {{E_2}} \right) = {3 \over 4}$$
<br><br>$$P\left( {{A \over {{E_1}}}} \right) = {{{}^{12}{C_2}} \ov... | mcq | jee-main-2021-online-16th-march-morning-shift | 7,651 |
1ktbbtchx | maths | probability | conditional-probability-and-multiplication-theorem | Let A and B be independent events such that P(A) = p, P(B) = 2p. The largest value of p, for which P (exactly one of A, B occurs) = $${5 \over 9}$$, is : | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$${4 \\over 9}$$"}, {"identifier": "D", "content": "$${5 \\over 12}$$"}] | ["D"] | null | P (Exactly one of A or B)<br><br>$$ = P\left( {A \cap \overline B } \right) + \left( {\overline A \cap B} \right) = {5 \over 9}$$<br><br>$$ = P(A)P(\overline B ) + P(\overline A )P(B) = {5 \over 9}$$<br><br>$$ \Rightarrow P(A)(1 - P(B)) + (1 - P(A))P(B) = {5 \over 9}$$<br><br>$$ \Rightarrow p(1 - 2p) + (1 - p)2p = {5 ... | mcq | jee-main-2021-online-26th-august-morning-shift | 7,652 |
1ktend64o | maths | probability | conditional-probability-and-multiplication-theorem | When a certain biased die is rolled, a particular face occurs with probability $${1 \over 6} - x$$ and its opposite face occurs with probability $${1 \over 6} + x$$. All other faces occur with probability $${1 \over 6}$$. Note that opposite faces sum to 7 in any die. If 0 < x < $${1 \over 6}$$, and the probabilit... | [{"identifier": "A", "content": "$${1 \\over 16}$$"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 9}$$"}, {"identifier": "D", "content": "$${1 \\over 12}$$"}] | ["B"] | null | Probability of obtaining total sum 7 = probability of getting opposite faces.<br><br>Probability of getting opposite faces<br><br>$$ = 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right]$$<br><br>$$ \Rightarrow 2\left[ {\le... | mcq | jee-main-2021-online-27th-august-morning-shift | 7,654 |
1ktit2n77 | maths | probability | conditional-probability-and-multiplication-theorem | An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second... | [] | null | 28 | I<sub>1</sub> = first unit is functioning<br><br>I<sub>2</sub> = second unit is functioning<br><br>P(I<sub>1</sub>) = 0.9, P(I<sub>2</sub>) = 0.8<br><br>P($$\overline {{I_1}} $$) = 0.1, P($$\overline {{I_2}} $$) = 0.2<br><br>$$P = {{0.8 \times 0.1} \over {0.1 \times 0.2 + 0.9 \times 0.2 + 0.1 \times 0.8}} = {8 \over {2... | integer | jee-main-2021-online-31st-august-morning-shift | 7,655 |
1l5aie7xz | maths | probability | conditional-probability-and-multiplication-theorem | <p>Let E<sub>1</sub> and E<sub>2</sub> be two events such that the conditional probabilities $$P({E_1}|{E_2}) = {1 \over 2}$$, $$P({E_2}|{E_1}) = {3 \over 4}$$ and $$P({E_1} \cap {E_2}) = {1 \over 8}$$. Then :</p> | [{"identifier": "A", "content": "$$P({E_1} \\cap {E_2}) = P({E_1})\\,.\\,P({E_2})$$"}, {"identifier": "B", "content": "$$P(E{'_1} \\cap E{'_2}) = P(E{'_1})\\,.\\,P(E{_2})$$"}, {"identifier": "C", "content": "$$P({E_1} \\cap E{'_2}) = P({E_1})\\,.\\,P({E_2})$$"}, {"identifier": "D", "content": "$$P(E{'_1} \\cap {E_2}) =... | ["C"] | null | <p>$$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}$$</p>
<p>$$P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}$$</p>
<p>$$P({E_1} \cap {E_2}) = {1 \over 8}$$</p>
<p>$$P(... | mcq | jee-main-2022-online-25th-june-morning-shift | 7,656 |
1l6f12wqy | maths | probability | conditional-probability-and-multiplication-theorem | <p>If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{3}, P(B)=\frac{1}{5}$$ and $$P(A \cup B)=\frac{1}{2}$$, then $$P\left(A \mid B^{\prime}\right)+P\left(B \mid A^{\prime}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{3}{4}$$"}, {"identifier": "B", "content": "$$\\frac{5}{8}$$"}, {"identifier": "C", "content": "$$\\frac{5}{4}$$"}, {"identifier": "D", "content": "$$\\frac{7}{8}$$"}] | ["B"] | null | <p>$$P(A) = {1 \over 3},\,P(B) = {1 \over 5}$$ and $$P\left( {A \cup B} \right) = {1 \over 2}$$</p>
<p>$$\therefore$$ $$P\left( {A \cap B} \right) = {1 \over 3} + {1 \over 5} - {1 \over 2} = {1 \over {30}}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7buiqub/55e764a3-4414-4eff-841a-7ce0120... | mcq | jee-main-2022-online-25th-july-evening-shift | 7,657 |
1l6m5ors7 | maths | probability | conditional-probability-and-multiplication-theorem | <p>Out of $$60 \%$$ female and $$40 \%$$ male candidates appearing in an exam, $$60 \%$$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :... | [{"identifier": "A", "content": "$$\\frac{2}{3}$$"}, {"identifier": "B", "content": "$$\\frac{11}{16}$$"}, {"identifier": "C", "content": "$$\\frac{23}{32}$$"}, {"identifier": "D", "content": "$$\\frac{13}{16}$$"}] | ["A"] | null | <p>P (Female) $$ = {{60} \over {100}} = {3 \over 5}$$</p>
<p>P (Male) $$ = {2 \over 5}$$</p>
<p>P (Female/Qualified) $$ = {{40} \over {60}} = {2 \over 3}$$</p>
<p>P (Male/qualified) $$ = {{20} \over {60}} = {1 \over 3}$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift | 7,658 |
1lgzyazug | maths | probability | conditional-probability-and-multiplication-theorem | <p>In a bolt factory, machines $$A, B$$ and $$C$$ manufacture respectively $$20 \%, 30 \%$$ and $$50 \%$$ of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufac... | [{"identifier": "A", "content": "$$\\frac{2}{7}$$"}, {"identifier": "B", "content": "$$\\frac{9}{28}$$"}, {"identifier": "C", "content": "$$\\frac{5}{14}$$"}, {"identifier": "D", "content": "$$\\frac{3}{7}$$"}] | ["C"] | null | Given : $P(A)=\frac{20}{100}=\frac{2}{10}$
<br/><br/>$$
P(B)=\frac{30}{100}=\frac{3}{10} $$
<br/><br/>$$ P(C)=\frac{50}{100}=\frac{5}{10}
$$
<br/><br/>Let $\mathrm{E} \rightarrow$ Event that the bolt is defective.
<br/><br/>$$
\text { So, } P(E / A)=\frac{3}{100}, $$
<br/><br/>$$P\left(\frac{E}{B}\right)=\frac{4}{100},... | mcq | jee-main-2023-online-8th-april-morning-shift | 7,661 |
lsamdflq | maths | probability | conditional-probability-and-multiplication-theorem | Let Ajay will not appear in JEE exam with probability $\mathrm{p}=\frac{2}{7}$, while both Ajay and Vijay will appear in the exam with probability $\mathrm{q}=\frac{1}{5}$. Then the probability, that Ajay will appear in the exam and Vijay will not appear is : | [{"identifier": "A", "content": "$\\frac{9}{35}$"}, {"identifier": "B", "content": "$\\frac{3}{35}$"}, {"identifier": "C", "content": "$\\frac{24}{35}$"}, {"identifier": "D", "content": "$\\frac{18}{35}$"}] | ["D"] | null | <p>We are given that the probability of Ajay not appearing in the JEE exam is $\mathrm{p}=\frac{2}{7}$, and the probability that both Ajay and Vijay will appear in the exam is $\mathrm{q}=\frac{1}{5}$.</p>
<p>We are asked to find the probability that Ajay will appear in the exam and Vijay will not. Let's denote this p... | mcq | jee-main-2024-online-1st-february-evening-shift | 7,662 |
lsblkyym | maths | probability | conditional-probability-and-multiplication-theorem | A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let <br/><br/>$a=P(X=3), b=P(X \geqslant 3)$ and $c=P(X \geqslant 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to __________. | [] | null | 12 | <p>To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\frac{b+c}{a}$.</p>
<p>Let's begin by defining each of the variables:</p>
<ul>
<li>$a = P(X=3)$: This is the probability that the first six appears on the third toss.</li>
<li>$b = P(X \... | integer | jee-main-2024-online-27th-january-morning-shift | 7,663 |
jaoe38c1lse58eei | maths | probability | conditional-probability-and-multiplication-theorem | <p>Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is</p> | [{"identifier": "A", "content": "$$\\frac{4}{25}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{25}$$\n"}, {"identifier": "D", "content": "$$\\frac{4}{75}$$"}] | ["D"] | null | <p>To solve this problem, we need to calculate the probability of two independent events occurring in succession: the first marble drawn is red, and the second marble drawn is white. Since the drawing is with replacement, the number of marbles of each color remains the same for both draws.</p>
<p>The total number of m... | mcq | jee-main-2024-online-31st-january-morning-shift | 7,664 |
jaoe38c1lseygjua | maths | probability | conditional-probability-and-multiplication-theorem | <p>A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is</p> | [{"identifier": "A", "content": "$$\\frac{5}{11}$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{6}$$\n"}, {"identifier": "D", "content": "$$\\frac{6}{11}$$"}] | ["A"] | null | <p>Required probability $$=$$</p>
<p>$$\begin{aligned}
& \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \\
& =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift | 7,665 |
m6SF5Xwnm5L4nJ2s | maths | probability | permutation-and-combination-based-problem | Four numbers are chosen at random (without replacement) from the set $$\left\{ {1,2,3,....20} \right\}.$$
<br/><br/><b>Statement - 1:</b> The probability that the chosen numbers when arranged in some order will form an AP is $${1 \over {85}}.$$
<p><b>Statement - 2:</b> If the four chosen numbers form an AP, then the ... | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false, Statement -2 is true."}, {"iden... | ["B"] | null | Four numbers can be chosen $${}^{20}{C_4}$$ ways.
<br><br>When common difference d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets
<br>So when d = 1 then 17 different AP's are possible with 4 numbers.
<br><br>Now let's create a table of all possible se... | mcq | aieee-2010 | 7,667 |
v0MDihlczrEKxer8 | maths | probability | permutation-and-combination-based-problem | If $$12$$ different balls are to be placed in $$3$$ identical boxes, then the probability that one of the boxes contains exactly $$3$$ balls is : | [{"identifier": "A", "content": "$$220{\\left( {{1 \\over 3}} \\right)^{12}}$$ "}, {"identifier": "B", "content": "$$22{\\left( {{1 \\over 3}} \\right)^{11}}$$"}, {"identifier": "C", "content": "$${{55} \\over 3}{\\left( {{2 \\over 3}} \\right)^{11}}$$ "}, {"identifier": "D", "content": "$$55{\\left( {{2 \\over 3}} \\r... | ["C"] | null | 1<sup>st</sup> ball can go any of the 3 boxes. So total choices for 1<sup>st</sup> ball = 3
<br><br>2<sup>nd</sup> ball can also go any of the 3 boxes. So total choices for 2<sup>nd</sup> ball = 3
<br>.
<br>.
<br>.
<br>.
<br>12<sup>th</sup> ball can go any of the 3 boxes. So total choices for 12<sup>th</sup> ball = 3
<... | mcq | jee-main-2015-offline | 7,668 |
hb12DaeYWQTsDI7quf0Na | maths | probability | permutation-and-combination-based-problem | From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :
| [{"identifier": "A", "content": "$${{21} \\over {220}}$$ "}, {"identifier": "B", "content": "$${{3} \\over {11}}$$ "}, {"identifier": "C", "content": "$${{1} \\over {11}}$$ "}, {"identifier": "D", "content": "$${{2} \\over {23}}$$ "}] | ["C"] | null | <p>The number of ways to form a committee having at least one woman is</p>
<p>$$ = {}^5{C_1} \times {}^{10}{C_3} + {}^5{C_2} \times {}^{10}{C_2} + {}^5{C_3} \times {}^{10}{C_1} + {}^5{C_4}$$</p>
<p>$$ = {{5!} \over {4!}} \times {{10!} \over {7! \times 3!}} + {{5!} \over {2!3!}} \times {{10!} \over {8!2!}} + {{5!} \over... | mcq | jee-main-2017-online-9th-april-morning-slot | 7,669 |
NAV5RgX2pFLPU0hYcWpBO | maths | probability | permutation-and-combination-based-problem | Two different families A and B are blessed with equal numbe of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $${1 \over {12}},$$ then the number of children ... | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["C"] | null | Let the number of children in each family be x.
<br><br>Thus the total number of children in both the families are 2x
<br><br>Now, it is given that 3 tickets are distributed amongst the children of these two families.
<br><br>Thus, the probability that all the three tickets go to the children in family B
<br><b... | mcq | jee-main-2018-online-16th-april-morning-slot | 7,670 |
izhPmlaiTmaj8SvcuP7k9k2k5khcyom | maths | probability | permutation-and-combination-based-problem | If 10 different balls are to be placed in 4 distinct
boxes at random, then the probability that two
of these boxes contain exactly 2 and 3 balls is : | [{"identifier": "A", "content": "$${{965} \\over {{2^{11}}}}$$"}, {"identifier": "B", "content": "$${{965} \\over {{2^{10}}}}$$"}, {"identifier": "C", "content": "$${{945} \\over {{2^{11}}}}$$"}, {"identifier": "D", "content": "$${{945} \\over {{2^{10}}}}$$"}] | ["D"] | null | Total ways of distribution = 4<sup>10</sup> = 2<sup>20</sup>
<br><br>Number of ways selecting two boxes out of four = <sup>4</sup>C<sub>2</sub>
<br><br>Then number of ways selecting 5 balls out of 10 = <sup>10</sup>C<sub>5</sub>
<br><br>Then no of ways of distributing 5 balls into two groups of 2 balls and 3 balls = <s... | mcq | jee-main-2020-online-9th-january-evening-slot | 7,673 |
D78VhXLXjBEt1DRB79jgy2xukf49fh2z | maths | probability | permutation-and-combination-based-problem | The probability that a randomly chosen 5-digit
number is made from exactly two digits is : | [{"identifier": "A", "content": "$${{150} \\over {{{10}^4}}}$$"}, {"identifier": "B", "content": "$${{134} \\over {{{10}^4}}}$$"}, {"identifier": "C", "content": "$${{121} \\over {{{10}^4}}}$$"}, {"identifier": "D", "content": "$${{135} \\over {{{10}^4}}}$$"}] | ["D"] | null | Sample space = 9 $$ \times $$ 10<sup>4</sup><br><br>Case - I<br><br>Out of exactly two digits selected one is zero then favourable cases = $${}^9{C_1}({2^4} - 1)$$<br><br>Case - II<br><br>Both selected digits are non-zero then favourable cases = $${}^9{C_2}({2^5} - 2)$$<br><br>Probability = $${{9({2^4} - 1) + {{9.8} \o... | mcq | jee-main-2020-online-3rd-september-evening-slot | 7,674 |
2gXd4NRSYBvLj2D1oajgy2xukfuvqwpi | maths | probability | permutation-and-combination-based-problem | Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is : | [{"identifier": "A", "content": "$${{10} \\over {99}}$$"}, {"identifier": "B", "content": "$${{5} \\over {33}}$$"}, {"identifier": "C", "content": "$${{15} \\over {101}}$$"}, {"identifier": "D", "content": "$${{5} \\over {101}}$$"}] | ["B"] | null | Out of 11 consecutive natural numbers either
6 even and 5 odd numbers or 5 even and 6
odd numbers.
<br><br>Let, E = Even
<br>O = Odd
<br><br><b>Case-1 :</b>
<br><br>E, O, E, O, E, O, E, O, E, O, E
<br><br>2b = a + c $$ \Rightarrow $$ Even
<br><br>$$ \Rightarrow $$ Both a and c should be either even or odd.
<br><br>P = ... | mcq | jee-main-2020-online-6th-september-morning-slot | 7,675 |
1krw0yts1 | maths | probability | permutation-and-combination-based-problem | Let 9 distinct balls be distributed among 4 boxes, B<sub>1</sub>, B<sub>2</sub>, B<sub>3</sub> and B<sub>4</sub>. If the probability than B<sub>3</sub> contains exactly 3 balls is $$k{\left( {{3 \over 4}} \right)^9}$$ then k lies in the set : | [{"identifier": "A", "content": "{x $$\\in$$ R : |x $$-$$ 3| < 1}"}, {"identifier": "B", "content": "{x $$\\in$$ R : |x $$-$$ 2| $$\\le$$ 1}"}, {"identifier": "C", "content": "{x $$\\in$$ R : |x $$-$$ 1| < 1}"}, {"identifier": "D", "content": "{x $$\\in$$ R : |x $$-$$ 5| $$\\le$$ 1}"}] | ["A"] | null | Required probability = $${{{}^9{C_3}{{.3}^6}} \over {{4^9}}}$$<br><br>$$ = {{{}^9{C_3}} \over {27}}.{\left( {{3 \over 4}} \right)^9}$$<br><br>$$ = {{28} \over 9}.{\left( {{3 \over 4}} \right)^9} \Rightarrow k = {{28} \over 9}$$<br><br>Which satisfies $$\left| {x - 3} \right| < 1$$ | mcq | jee-main-2021-online-25th-july-morning-shift | 7,681 |
1ktk4uzqz | maths | probability | permutation-and-combination-based-problem | Let S = {1, 2, 3, 4, 5, 6}. Then the probability that a randomly chosen onto function g from S to S satisfies g(3) = 2g(1) is : | [{"identifier": "A", "content": "$${1 \\over {10}}$$"}, {"identifier": "B", "content": "$${1 \\over {15}}$$"}, {"identifier": "C", "content": "$${1 \\over {5}}$$"}, {"identifier": "D", "content": "$${1 \\over {30}}$$"}] | ["A"] | null | g(3) = 2g(1) can be defined in 3 ways<br><br>number of onto functions in this condition = 3 $$\times$$ 4!<br><br>Total number of onto functions = 6!<br><br>Required probability = $${{3 \times 4!} \over {6!}} = {1 \over {10}}$$ | mcq | jee-main-2021-online-31st-august-evening-shift | 7,682 |
1kto2k0k0 | maths | probability | permutation-and-combination-based-problem | Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :<br/><br/><img src="data:image/png;base64,UklGRsgPAABXRUJQVlA4ILwPAAAw1ACdASoAA8sBP4G21WS2LTenINF7OvAwCWlLUXMT93fZNRcQq8/yf42HV04Op8xQDpl+Vojfnw/PvNf8+1hXFHdD2ge2X918KbeHt4WjD0Un//6sTuqxHG4Rp02vZ0F7Yc8nhR5... | [{"identifier": "A", "content": "$${2 \\over 7}$$"}, {"identifier": "B", "content": "$${1 \\over 18}$$"}, {"identifier": "C", "content": "$${1 \\over 7}$$"}, {"identifier": "D", "content": "$${1 \\over 9}$$"}] | ["B"] | null | Total ways of choosing square = $${}^{64}{C_2}$$<br><br>$$ = {{64 \times 63} \over {2 \times 1}} = 32 \times 63$$<br><br>ways of choosing two squares having common side = 2 (7 $$\times$$ 8) = 112<br><br>Required probability $$ = {{112} \over {32 \times 63}} = {{16} \over {32 \times 9}} = {1 \over {18}}$$.<br><br>Ans. (... | mcq | jee-main-2021-online-1st-september-evening-shift | 7,683 |
1l544hp5b | maths | probability | permutation-and-combination-based-problem | <p>The probability that a randomly chosen 2 $$\times$$ 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :</p> | [{"identifier": "A", "content": "$${{133} \\over {{{10}^4}}}$$"}, {"identifier": "B", "content": "$${{18} \\over {{{10}^3}}}$$"}, {"identifier": "C", "content": "$${{19} \\over {{{10}^3}}}$$"}, {"identifier": "D", "content": "$${{271} \\over {{{10}^4}}}$$"}] | ["C"] | null | <p>First 10 prime numbers are</p>
<p>={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}</p>
<p>Let A is a 2 $$\times$$ 2 matrix,</p>
<p>$$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$</p>
<p>Given that matrix A is singular.</p>
<p>$$\therefore$$ | A | = 0</p>
<p>$$ \Rightarrow \left| {\matrix{
a & b \cr
... | mcq | jee-main-2022-online-29th-june-morning-shift | 7,684 |
1l55iov0n | maths | probability | permutation-and-combination-based-problem | <p>The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b) $$-$$ f(c) = f(d) is :</p> | [{"identifier": "A", "content": "$${1 \\over {24}}$$"}, {"identifier": "B", "content": "$${1 \\over {40}}$$"}, {"identifier": "C", "content": "$${1 \\over {30}}$$"}, {"identifier": "D", "content": "$${1 \\over {20}}$$"}] | ["D"] | null | Number of one-one function from $\{a, b, c, d\}$ to set $\{1,2,3,4,5\}$ is ${ }^{5} P_{4}=120 n(s)$.
<br/><br/>
The required possible set of value (f(a), $f(b), f(c), f(d))$ such that $f(a)+2 f(b)-f(c)=f(d)$ are $(5,3,2,1),(5,1,2,3),(4,1,3,5),(3,1,4,5)$, $(5,4,3,2)$ and $(3,4,5,2)$
<br/><br/>
$\therefore n(E)=6$
<br/><... | mcq | jee-main-2022-online-28th-june-evening-shift | 7,685 |
1l57onrrd | maths | probability | permutation-and-combination-based-problem | <p>Five numbers $${x_1},{x_2},{x_3},{x_4},{x_5}$$ are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order $$({x_1} < {x_2} < {x_3} < {x_4} < {x_5})$$. The probability that $${x_2} = 7$$ and $${x_4} = 11$$ is :</p> | [{"identifier": "A", "content": "$${1 \\over {136}}$$"}, {"identifier": "B", "content": "$${1 \\over {72}}$$"}, {"identifier": "C", "content": "$${1 \\over {68}}$$"}, {"identifier": "D", "content": "$${1 \\over {34}}$$"}] | ["C"] | null | No. of ways to select and arrange $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5$ from $1,2,3$.......18<br/><br/>
$$
\begin{aligned}
& \mathrm{n}(\mathrm{s})={ }^{18} \mathrm{C}_5 \\\\
& \begin{array}{lllll}
x_1 & \underset{7}{\left(x_2\right)} & x_3 & \underset{11}{\left(x_4\right)} & x_5
\end{a... | mcq | jee-main-2022-online-27th-june-morning-shift | 7,686 |
1l58h8xsg | maths | probability | permutation-and-combination-based-problem | <p>If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to _______________.</p> | [] | null | 33 | <p>Total number of numbers from given</p>
<p>Condition = n(s) = 2<sup>6</sup>.</p>
<p>Every required number is of the form</p>
<p>A = 7 . (10<sup>a<sub>1</sub></sup> + 10<sup>a<sub>2</sub></sup> + 10<sup>a<sub>3</sub></sup> + .......) + 111111</p>
<p>Here 111111 is always divisible by 21.</p>
<p>$$\therefore$$ If A is ... | integer | jee-main-2022-online-26th-june-evening-shift | 7,687 |
1ldsuemqy | maths | probability | permutation-and-combination-based-problem | <p>Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{6}$$"}, {"identifier": "B", "content": "$$\\frac{2}{15}$$"}, {"identifier": "C", "content": "$$\\frac{5}{24}$$"}, {"identifier": "D", "content": "0.08"}] | ["D"] | null | Required probability $=1-\frac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}$
<br/><br/>Taking $\mathrm{D}_{(15)}$ as $\frac{15 !}{e}$
<br/><br/>$\mathrm{D}_{(14)}$ as $\frac{14 \text { ! }}{e}$
<br/><br/>$\mathrm{D}_{(13)}$ as $\frac{13 \text { ! }}{e}$
<br/><br/>$$
\text { We get, } \text { Requir... | mcq | jee-main-2023-online-29th-january-morning-shift | 7,688 |
jaoe38c1lscn159w | maths | probability | permutation-and-combination-based-problem | <p>An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :</p> | [{"identifier": "A", "content": "$$\\frac{3}{256}$$"}, {"identifier": "B", "content": "$$\\frac{5}{256}$$"}, {"identifier": "C", "content": "$$\\frac{3}{715}$$"}, {"identifier": "D", "content": "$$\\frac{5}{715}$$"}] | ["C"] | null | <p>$$\frac{{ }^6 \mathrm{C}_4}{{ }^{15} \mathrm{C}_4} \times \frac{{ }^9 \mathrm{C}_4}{{ }^{11} \mathrm{C}_4}=\frac{3}{715}$$</p>
<p>Hence option (3) is correct.</p> | mcq | jee-main-2024-online-27th-january-evening-shift | 7,689 |
jaoe38c1lsd3atnw | maths | probability | permutation-and-combination-based-problem | <p>A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is</p> | [{"identifier": "A", "content": "$$\\frac{1}{9}$$"}, {"identifier": "B", "content": "$$\\frac{2}{9}$$"}, {"identifier": "C", "content": "$$\\frac{1}{27}$$"}, {"identifier": "D", "content": "$$\\frac{2}{27}$$"}] | ["B"] | null | <p>To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).</p>
<p>Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as $$ P(T) = p $$ and the probability of getting a head as $$ P(H) = 2p $$.</p>... | mcq | jee-main-2024-online-31st-january-evening-shift | 7,690 |
lv9s209z | maths | probability | permutation-and-combination-based-problem | <p>The coefficients $$\mathrm{a}, \mathrm{b}, \mathrm{c}$$ in the quadratic equation $$\mathrm{a} x^2+\mathrm{bx}+\mathrm{c}=0$$ are from the set $$\{1,2,3,4,5,6\}$$. If the probability of this equation having one real root bigger than the other is p, then 216p equals :</p> | [{"identifier": "A", "content": "38"}, {"identifier": "B", "content": "76"}, {"identifier": "C", "content": "57"}, {"identifier": "D", "content": "19"}] | ["A"] | null | <p>Equation is $$a x^2+b x+c=0$$</p>
<p>$$\mathrm{D}>0$$ [for roots to be real & distinct]</p>
<p>$$\Rightarrow b^2-4 a c>0$$</p>
<p>For $$b<2$$ no value of $$a$$ & $$c$$ are possible</p>
<p>$$\begin{aligned}
& \text { For } b=3 \Rightarrow a c<\frac{9}{4} \\
& (a, c) \in\{(1,1),(1,2),(2,1)\} \Rightarrow 3 \text { case... | mcq | jee-main-2024-online-5th-april-evening-shift | 7,691 |
uiR8NU8AS1zaGHd3 | maths | probability | probability-distribution-of-a-random-variable | A random variable $$X$$ has Poisson distribution with mean $$2$$.
<br/>Then $$P\left( {X > 1.5} \right)$$ equals : | [{"identifier": "A", "content": "$${2 \\over {{e^2}}}$$ "}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$1 - {3 \\over {{e^2}}}$$ "}, {"identifier": "D", "content": "$${3 \\over {{e^2}}}$$ "}] | ["C"] | null | <p>In a position distribution,</p>
<p>$$P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}$$ ($$\lambda$$ = mean).</p>
<p>Now, $$P(X = r > 1.5) = P(2) + P(3) + $$ ..... $$\infty$$</p>
<p>$$ = 1 - \{ P(0) + P(1)\} $$</p>
<p>$$ = 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}... | mcq | aieee-2005 | 7,692 |
ZjpgMVecKoOHSSxa | maths | probability | probability-distribution-of-a-random-variable | At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of $$5$$ phone calls during $$10$$ minute time intervals. The probability that there is at the most one phone call during a $$10$$-minute time period is : | [{"identifier": "A", "content": "$${6 \\over {{5^e}}}$$ "}, {"identifier": "B", "content": "$${5 \\over 6}$$ "}, {"identifier": "C", "content": "$${6 \\over 55}$$"}, {"identifier": "D", "content": "$${6 \\over {{e^5}}}$$ "}] | ["D"] | null | <p>Required probability</p>
<p>$$ = P(X = 0) + P(X = 1)$$</p>
<p>$$ = {{{e^{ - 5}}} \over {0!}}{5^0} + {{{e^{ - 5}}} \over {1!}}{5^1}$$</p>
<p>$$ = {e^{ - 5}} + 5{e^{ - 5}}$$</p>
<p>$$ = {6 \over {{e^5}}}$$</p> | mcq | aieee-2006 | 7,693 |
P6xO9hto9apIDYTTsa3rsa0w2w9jxayoap3 | maths | probability | probability-distribution-of-a-random-variable | A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins
Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the
expected gain/loss (in Rs.) of the person is : | [{"identifier": "A", "content": "$${1 \\over 4}$$ loss"}, {"identifier": "B", "content": "$${1 \\over 2}$$ gain"}, {"identifier": "C", "content": "$${1 \\over 2}$$ loss"}, {"identifier": "D", "content": "2 gain"}] | ["C"] | null | When two dice are thrown then sample space will {(1, 1), (2, 2) ....... (6, 6)} contain total 36 elements number of cases.<br><br>
Then the expectation will be $${6 \over {36}} \times 15 \times {4 \over {36}} \times 12 - {{26} \over {36}} \times 6$$<br><br>
$${{90 + 48 - 156} \over {36}} = - {1 \over 2}$$ = $$ {1 \ove... | mcq | jee-main-2019-online-12th-april-evening-slot | 7,694 |
FxEzwiF0BbebQgJ2yf7k9k2k5e2z1s8 | maths | probability | probability-distribution-of-a-random-variable | An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value of k when k
consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then the expected
value of X, is :
| [{"identifier": "A", "content": "$$ - {3 \\over {16}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${3 \\over {16}}$$"}] | ["C"] | null | Number of ways 3 consecutive heads can appers
<br><br>(1) HHHT_
<br><br>(2) _THHH
<br><br>(3) THHHT
<br><br>$$ \therefore $$ Probablity of getting 3 consecutive heads
<br><br>= $${2 \over {32}}$$ + $${2 \over {32}}$$ + $${1 \over {32}}$$ = $${5 \over {32}}$$
<br><br>Number of ways 4 consecutive heads can appers
<br><br... | mcq | jee-main-2020-online-7th-january-morning-slot | 7,695 |
F1meBFGPVyGyNy1nKL7k9k2k5kh288t | maths | probability | probability-distribution-of-a-random-variable | A random variable X has the following
probability distribution :<br/><br/>
<style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width:100%}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.... | [{"identifier": "A", "content": "$${1 \\over {6}}$$"}, {"identifier": "B", "content": "$${7 \\over {12}}$$"}, {"identifier": "C", "content": "$${1 \\over {36}}$$"}, {"identifier": "D", "content": "$${23 \\over {36}}$$"}] | ["D"] | null | $$\sum\limits_{i = 1}^5 {P(X)} $$ = 1
<br><br>$$ \Rightarrow $$ K<sup>2</sup>
+ 2K + K + 2K + 5K<sup>2</sup>
= 1
<br><br>$$ \Rightarrow $$ 6K<sup>2</sup>
+ 5K – 1 = 0
<br><br>$$ \Rightarrow $$ (6K - 1)(k + 1) = 0
<br><br>$$ \Rightarrow $$ K = $${1 \over 6}$$ and K = -1(rejected)
<br><br>$$ \therefore $$ P(X $$ > ... | mcq | jee-main-2020-online-9th-january-evening-slot | 7,696 |
1ktgohohi | maths | probability | probability-distribution-of-a-random-variable | The probability distribution of random variable X is given by :<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{bor... | [] | null | 30 | $$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$$<br><br>$$ \Rightarrow k = {1 \over 9}$$<br><br>Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$<br><br>$$ \Rightarrow p = {2 \over 3}$$<br... | integer | jee-main-2021-online-27th-august-evening-shift | 7,698 |
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