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2. During the transportation of a box containing 21 standard and 10 non-standard parts, one part was lost, and it is unknown which one. A part randomly extracted from the box (after transportation) turned out to be standard. Find the probability that the lost part was: a) a standard part; b) a non-standard part. | Solution. a) The extracted standard part obviously could not have been lost; any of the other 30 parts could have been lost $(21+10-1=30)$, and among them, 20 were standard $(21-1=20)$. The probability that a standard part was lost, $P=20 / 30=2 / 3$.
b) Among the 30 parts, each of which could have been lost, 10 were ... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,284 |
11. In a box, there are 10 identical parts marked with numbers 1, 2, ..., 10. Six parts are randomly drawn. Find the probability that among the drawn parts there will be: a) part No. 1; b) parts No. 1 and No. 2. | Solution. a) The total number of possible elementary outcomes of the experiment is equal to the number of ways in which six parts can be drawn from ten, i.e., $C_{10}^{6}$.
Let's find the number of outcomes favorable to the event we are interested in: among the six selected parts, there is part 1, and consequently, th... | 0.6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,285 |
17. In a batch of $N$ parts, there are $n$ standard ones. $m$ parts are randomly selected. Find the probability that among the selected parts, exactly $k$ are standard. | Solution. The total number of possible elementary outcomes of the trial is equal to the number of ways to extract $m$ parts from $N$ parts, i.e., $C_{N}^{m}$ - the number of combinations of $N$ elements taken $m$ at a time.
Let's calculate the number of outcomes favorable to the event of interest (exactly $k$ standard... | C_{n}^{k}\cdotC_{N-n}^{-k}/C_{N}^{} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,286 |
23. The quality control department detected five defective books in a batch of 100 randomly selected books. Find the relative frequency of the appearance of defective books. | Solution. The relative frequency of event $\boldsymbol{A}$ (the appearance of defective books) is equal to the ratio of the number of trials in which event $A$ occurred to the total number of trials conducted: $W(A)=5 / 100=0.05$. | 0.05 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,287 |
41. A signal detector receives signals from two devices, and the arrival of each signal is equally likely at any moment within a time interval of duration $T$. The moments of signal arrival are independent of each other. The detector triggers if the difference between the moments of signal arrival is less than $t(t<T)$... | Solution. Let the moments of signal arrivals from the first and second devices be denoted by $x$ and $y$, respectively. According to the problem's condition, the following double inequalities must be satisfied: $0 \leq x$ and $x - t \leq y \leq x$, or equivalently,
$$
\begin{array}{lll}
y \leq x \\
y > x - t & \text {... | \frac{(2T-)}{T^2} | Other | math-word-problem | Yes | Yes | olympiads | false | 32,291 |
46. On a library shelf, 15 textbooks are randomly arranged, with five of them in hardcover. The librarian randomly takes three textbooks. Find the probability that at least one of the taken textbooks will be in hardcover (event $A$). | Solution. First method. The requirement - at least one of the three taken textbooks is in a hardcover - will be fulfilled if any of the following three mutually exclusive events occur: $B$ - one textbook is in a hardcover, $C$ - two textbooks are in a hardcover, $D$ - three textbooks are in a hardcover.
The event of i... | \frac{67}{91} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,292 |
47. In a box, there are 10 parts, four of which are painted. A worker randomly took three parts. Find the probability that at least one of the taken parts is painted.
48. Prove that if event $A$ implies event $B$, then $P(B) \geqslant P(A)$. | Solution. Event $B$ can be represented as the sum of non-overlapping events $A$ and $\bar{A} B$:
$$
B=A \div \bar{A} B
$$
By the theorem of addition of probabilities of non-overlapping events, we get
$$
P(B)=P(A+\bar{A} B)=P(A)+P(\bar{A} B)
$$
Since $P(\bar{A} B) \geqslant 0$, then $P(B) \geqslant P(A)$. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,293 |
49. The probabilities of occurrence of each of two independent events $A_{1}$ and $A_{2}$ are $p_{1}$ and $p_{2}$, respectively. Find the probability of the occurrence of only one of these events. | Solution. Let's introduce the notation for events: $B_{1}$ - only event $A_{1}$ occurs; $B_{2}$ - only event $A_{2}$ occurs.
The occurrence of event $B_{1}$ is equivalent to the occurrence of event $A_{1} \bar{A}_{2}$ (the first event occurs and the second does not), i.e., $B_{1}=A_{1} \bar{A}_{2}$. The occurrence of ... | p_{1}q_{2}+q_{1}p_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,294 |
64. In the reading room, there are six textbooks on probability theory, three of which are in hardcover. The librarian randomly took two textbooks. Find the probability that both textbooks will be in hardcover. | Solution. Let's introduce the notation for events: $A$ - the first textbook has a cover, $B$ - the second textbook has a cover. The probability that the first textbook has a cover, $P(A)=3 / 6=1 / 2$.
The probability that the second textbook has a cover, given that the first taken textbook was in a cover, i.e., the co... | 0.2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,295 |
66. In the workshop, seven men and three women work. By their employee numbers, three people are randomly selected. Find the probability that all the selected individuals will be men. | Solution. Let's introduce the designations of events: $A$ - the first selected is a man; $B$ - the second selected is a man; $C$ - the third selected is a man. The probability that the first selected is a man, $P(A)=7 / 10$.
The probability that the second selected is a man, given that the first selected was a man, i.... | \frac{7}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,296 |
74. Find the probability $P(\bar{A} \bar{B})$ given the following probabilities:
$$
P(A)=a, P(B)=b . \quad P(A+B)=c
$$ | Solution. Using the identity $P(\bar{B})=P(A \bar{B})+P(\bar{A} \bar{B})$, we find $P(\bar{A} \bar{B}):$
$$
P(\bar{A} \bar{B})=P(\bar{B})-P(A \bar{B})=(1-b)-P(A \bar{B})
$$
Substituting into the last equality $P(A \bar{B})=c-b$ (see problem 73), we get
$$
P(\bar{A} \bar{B})=1-b-(c-b)=1-c
$$ | 1- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,298 |
75. The occurrence of event $A B$ necessarily leads to the occurrence of event $C$. Prove that $P(A)+P(B)$ $-P(C) \leqslant 1$. | Solution. By the condition, the occurrence of event $AB$ implies the occurrence of event $C$, therefore (see problem 48)
$$
P(C) \geqslant P(A B)
$$
Using the identities
$$
\begin{gathered}
P(A)=P(A B)+P(A \bar{B}), P(B)=P(A B)+P(\bar{A} B) \\
P(A B)+P(A \bar{B})+P(\overline{A B})=1-P(\bar{A} B)
\end{gathered}
$$
a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,299 |
76. Prove that
$$
P_{A}(B) \geqslant 1-\frac{P(\bar{B})}{P(A)}
$$
It is assumed that $P(A)>0$. | Solution. By the remark to problem 75, the inequality holds
$$
P(A)+P(B)-P(A B)<1
$$
We will use the identities
$$
P(A B)=P(A) \cdot P_{A}(B), \quad P(B)=1-P(\bar{B}) .
$$
Substituting (**) into (*), we get
$$
P(A)+1-P(\bar{B})-P(A) \cdot P_{A}(B)<1 \text {, }
$$
or
$$
P(A) \cdot P_{A}(B) \geqslant P(A)-P(\bar{B... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,300 |
77. The occurrence of events $A B C$ necessarily leads to the occurrence of event $D$. Prove that
$$
P(A)+P(B)+P(C)-P(D) \leqslant 2
$$ | Solution. According to the condition, the occurrence of the event $A B C$ necessarily leads to the occurrence of event $D$, therefore (see problem 48), $\boldsymbol{P}(D) \geqslant P(A B C)$. Thus, if the inequality
$$
P(A)+P(B)+P(C)-P(A B C)<2
$$
is proven, then the inequality specified in the problem condition will... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,301 |
78. To state the addition theorem of probabilities for three joint events:
$$
\begin{aligned}
& P(A+B+C)=P(A)+P(B)+P(C)- \\
& -P(A B)-P(A C)-P(B C)+P(A B C)
\end{aligned}
$$
It is assumed that the addition theorem for two joint events has already been proven:
$$
P\left(A_{1}+A_{2}\right)=P\left(A_{1}\right)+P\left(A... | Solution. We reduce the sum of three events to the sum of two events:
$$
A+B+C=(A+B)+C
$$
We use the theorem of addition of probabilities of two events:
$$
\begin{aligned}
& P(A+B+C)=P[(A+B)+C]= \\
= & P(A+B)+P(C)-P[(A+B) C]= \\
= & P(A+B)+P(C)-P[(A C)+(B C)] .
\end{aligned}
$$
We apply the theorem of addition of p... | \frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 32,302 |
84. The probability of successfully performing an exercise for each of two athletes is 0.5. The athletes perform the exercise in turns, with each making two attempts. The one who performs the exercise first wins the prize. Find the probability of the athletes winning the prize. | Solution. For the prize to be awarded, it is sufficient for at least one of the four attempts to be successful. The probability of a successful attempt is $p=0.5$, and of an unsuccessful attempt is $q=1-0.5=0.5$. The desired probability is
$$
P=1-q^{4}=1-0.5^{4}=0.9375
$$ | 0.9375 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,303 |
86. The probability of at least one hit by a shooter on a target in three shots is 0.875. Find the probability of a hit in one shot. | Solution. The probability of hitting the target at least once in three shots (event $A$) is
$$
P(A)=1-q^{3} \text {, }
$$
where $q$ is the probability of a miss.
According to the condition, $P(A)=0.875$. Therefore,
$$
0.875=1-q^{3}, \text { or } q^{3}=1-0.875=0.125 \text {. }
$$
Hence, $q=\sqrt[3]{0.125}=0.5$.
Th... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,304 |
89. A white ball is placed into an urn containing two balls, after which one ball is randomly drawn from it. Find the probability that the drawn ball will be white, if all possible assumptions about the initial composition of the balls (by color) are equally likely. | Solution. Let $A$ be the event of drawing a white ball. The following hypotheses about the original composition of the balls are possible: $B_{1}$ - no white balls, $B_{2}$ - one white ball, $B_{3}$ - two white balls.
Since there are three hypotheses, and by condition they are equally probable, and the sum of the prob... | \frac{2}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,305 |
105. There are three batches of parts, each containing 20 parts. The number of standard parts in the first, second, and third batches is $20, 15, 10$, respectively. A part is randomly selected from a randomly chosen batch and turns out to be standard. The part is returned to the batch, and a second part is randomly sel... | Solution. Let $A$ be the event that in each of the two trials (with replacement) a standard part was drawn.
We can make three assumptions (hypotheses): $B_{1}$ - parts were drawn from the first batch; $B_{2}$ - parts were drawn from the second batch; $B_{3}$ - parts were drawn from the third batch.
The parts were dra... | \frac{4}{29} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,306 |
106. A battery of three guns fired a salvo, with two shells hitting the target. Find the probability that the first gun hit the target, given that the probabilities of hitting the target with the first, second, and third guns are respectively $p_{1}=0.4, p_{2}=0.3, p_{3}=0.5$. | Solution. Let $A$ be the event that two guns hit the target. We make two assumptions (hypotheses): $B_{1}$ - the first gun hit the target; $B_{2}$ - the first gun missed the target.
According to the condition, $P\left(B_{1}\right)=0.4$; therefore (event $B_{2}$ is the opposite of event $B_{1}$),
$$
P\left(B_{2}\right... | \frac{20}{29} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,307 |
108. Two out of three independently operating elements of a computing device have failed. Find the probability that the first and second elements have failed, if the probabilities of failure of the first, second, and third elements are 0.2; 0.4 and 0.3, respectively. | Solution. Let $A$ be the event that two elements have failed. The following assumptions (hypotheses) can be made:
$B_{1}$ - the first and second elements have failed, while the third element is intact, and (since the elements operate independently, the multiplication theorem applies)
$$
P\left(B_{1}\right)=p_{1} \cdo... | 0.3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,308 |
110. Two equally skilled chess players are playing chess. What is more likely: to win two out of four games or three out of six (draws are not considered)? | Solution. The chess players are of equal strength, so the probability of winning $p=1 / 2$; consequently, the probability of losing $q$ is also $1 / 2$. Since the probability of winning is constant in all games and the order in which the games are won does not matter, the Bernoulli formula can be applied.
Let's find t... | P_{4}(2)>P_{6}(3) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,309 |
119. Find the probability that event $A$ will occur exactly 70 times in 243 trials, if the probability of this event occurring in each trial is 0.25. | Solution. According to the condition, $n=243 ; k=70 ; p=0.25 ; q=0.75$. Since $n=243$ is a sufficiently large number, we will use the local theorem of Laplace:
$$
P_{n}(k)=\frac{1}{\sqrt{n p q}} \varphi(x)
$$
where $x=(k-n p) / \sqrt{n p q}$.
Let's find the value of $x$:
$$
x=\frac{k-n p}{\sqrt{n p q}}=\frac{70-243... | 0.0231 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,310 |
120. Find the probability that event $A$ will occur 1400 times in 2400 trials, if the probability of this event occurring in each trial is 0.6. | Solution. Since $\boldsymbol{n}$ is large, we will use the local Laplace theorem:
$$
P_{n}(k)=\frac{1}{\sqrt{n q p}} \varphi(x)
$$
Let's calculate $x:$
$$
x=\frac{k-n p}{\sqrt{n p q}}=\frac{1400-2400 \cdot 0,6}{\sqrt{2400 \cdot 0,6 \cdot 0,4}}=-\frac{40}{24}=-1,67
$$
The function $\varphi(x)=\frac{1}{\sqrt{2 \pi}} ... | 0.0041 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,311 |
125. The probability of an event occurring in each of 100 independent trials is constant and equal to \( p = 0.8 \). Find the probability that the event will occur: a) at least 75 times and no more than 90 times; b) at least 75 times; c) no more than 74 times. | Solution. We will use the integral Laplace theorem:
$$
P_{n}\left(k_{1}, k_{2}\right)=\Phi\left(x^{\prime \prime}\right)-\Phi\left(x^{\prime}\right),
$$
where $\Phi(x)$ is the Laplace function,
$$
x^{\prime}=\left(k_{1}-n p\right) / \sqrt{\overline{n p q}}, \quad x^{\prime \prime}=\left(k_{2}-n p\right) / \sqrt{n p ... | 0.8882,0.8944,0.1056 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,312 |
129. The probability of an event occurring in each of the independent trials is 0.8. How many trials need to be conducted to expect with a probability of 0.9 that the event will occur at least 75 times? | Solution. According to the condition, $p=0.8 ; q=0.2 ; k_{i}=75 ; k_{2}=n$; $P_{n}=(75, n)=0.9$.
Let's use the integral theorem of Laplace:
$$
P_{n}\left(k_{1} ; n\right)=\Phi\left(x^{\prime \prime}\right)-\Phi\left(x^{\prime}\right)=\Phi\left[\frac{k_{2}-n p}{\sqrt{n p q}}\right]-\Phi\left[\frac{k_{1}-n p}{\sqrt{n p... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,313 |
131. The probability of an event occurring in each of 625 independent trials is 0.8. Find the probability that the relative frequency of the event's occurrence deviates from its probability in absolute value by no more than 0.04. | Solution. According to the condition, $n=625 ; p=0.8 ; q=0.2 ; \varepsilon=0.04$. It is required to find the probability $P(|m / 625-0.8|<0.04)$. We will use the formula
$$
P\left(\left|\frac{m}{n}-p\right|<\varepsilon\right)=2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
$$
We have
$$
P\left(\left|\frac{m}{625... | 0.9876 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,314 |
136. How many times should a die be thrown so that the probability of the inequality
$$
|m / n-1 / 6| \leqslant 0.01
$$
is not less than the probability of the opposite inequality, where $m$ is the number of times one dot appears in $n$ throws of a die? | Solution. We will use the formula
$$
P\left(\left|\frac{m}{n}-p\right|<\varepsilon\right)=1-2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
$$
According to the condition, the inequality should hold
$$
2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right) \geqslant 1-2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\rig... | 632 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 32,315 |
139. The probability of an event occurring in each of 400 independent trials is 0.8. Find a positive number $\varepsilon$ such that with a probability of 0.99, the absolute value of the deviation of the relative frequency of the event's occurrence from its probability of 0.8 does not exceed $\varepsilon$. | Solution. According to the condition, $n=400, p=0.8, q=0.2$. Therefore,
$$
2 \Phi(\varepsilon \sqrt{400 /(0.8 \cdot 0.2)})=0.99 \quad \text { or } \quad \Phi(50 \varepsilon)=0.495 .
$$
From the table in Appendix 2, we find $\Phi(2.57)=0.495$, which means $50 \varepsilon=$ $=2.57$. From this, the required value $\vare... | 0.05 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,316 |
142. The quality control department checks 900 parts for standardization. The probability that a part is standard is 0.9. Find, with a probability of 0.95, the boundaries within which the number $m$ of standard parts among those checked will be contained. | The solution is as follows. Given $n=900, p=0.9, q=0.1$. Therefore,
$$
2 \Phi(\varepsilon \sqrt{900 /(0.9 \cdot 0.1)})=0.95, \text { or } \Phi(100 \varepsilon)=0.475
$$
From the table in Appendix 2, we find $\Phi(1.96)=0.475$, so $100 \varepsilon=1.96$. Hence, $\varepsilon \approx 0.02$.
Thus, with a probability of ... | 792\leqslant\leqslant828 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,317 |
145. Each of the 15 elements of a certain device is tested. The probability that an element will withstand the test is 0.9. Find the most probable number of elements that will withstand the test. | Solution. According to the condition, $n=15, p=0.9, q=0.1$. Let's find the most probable number $k_{0}$ from the double inequality
$$
n p-q<k_{0}<n p+p
$$
Substituting the data from the problem, we get
$$
15 \cdot 0.9-0.1 \leqslant k_{0}<15 \cdot 0.9+0.9, \text{ or } 13.5 \leqslant k_{0}<14.4 \text{. }
$$
Since $k_... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,318 |
147. A goods expert examines 24 samples of goods. The probability that each sample will be deemed suitable for sale is 0.6. Find the most probable number of samples that the goods expert will consider suitable for sale. | Solution. According to the condition, $n=24 ; p=0.6 ; q=0.4$. Let's find the most probable number of saleable samples from the double inequality $n p-q<k_{0}<n p+p$. Substituting the data from the problem, we get
$$
24 \cdot 0.6-0.4<k_{0}<24 \cdot 0.6+0.6 \text {, or } 14<k_{0}<15 \text {. }
$$
Since $n p-q=14$ is an... | 1415 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,319 |
149. Two equally matched opponents are playing chess. Find the most probable number of wins for any chess player if $2 N$ decisive (without draws) games will be played. | Solution. It is known that if the product of the number of trials $\boldsymbol{n}$ and the probability $p$ of the event occurring in one trial is an integer, then the most probable number is
$$
k_{0}=n p .
$$
In the problem at hand, the number of trials $n$ is equal to the number of games played $2 N$; the probabilit... | N | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,320 |
150. Two shooters shoot at a target. The probability of a miss with one shot for the first shooter is 0.2, and for the second shooter it is 0.4. Find the most probable number of volleys in which there will be no hits on the target, if the shooters will make 25 volleys. | Solution. Misses by the shooters are independent events, so the multiplication theorem of probabilities of independent events applies. The probability that both shooters will miss in one volley, $p=0.2 \cdot 0.4=0.08$.
Since the product $n p=25 \cdot 0.08=2$ is an integer, the most probable number of volleys in which ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,321 |
152. How many independent trials with a probability of the event occurring in each trial equal to 0.4 are needed so that the most probable number of occurrences of the event in these trials is 25? | Solution. According to the condition, $k_{0}=25 ; p=0.4 ; q=0.6$. We will use the double inequality
$$
n p-q \leqslant k_{0}25
$$
From the first inequality of the system, we find $n24.6 / 0.4=61.5$.
Thus, the required number of trials must satisfy the double inequality $62 \leqslant n<64$. | 62\leqslantn<64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,322 |
155. What is the probability \( p \) of an event occurring in each of 49 independent trials, if the most probable number of occurrences of the event in these trials is 30? | Solution. According to the condition, $n=49, k_{0}=30$. We will use the double inequality $n p-q \leqslant k_{0}30, \quad 49 p-(1-p)0,6$. From the second inequality of the system, we find $p \leqslant 0,62$.
Thus, the required probability must satisfy the double inequality $0,6<p \leq 0,62$. | 0.6<p\leq0.62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,323 |
157. The battery fired six shots at the object. The probability of hitting the object with one shot is 0.3. Find: a) the most probable number of hits; b) the probability of the most probable number of hits; c) the probability that the object will be destroyed, if for this at least two hits are required. | Solution. According to the condition, $n=6 ; p=0.3 ; q=0.7$. a) Let's find the most probable number of hits using the formula
$$
n p-q<k_{0}<n p+p
$$
Substituting the data from the problem, we get
$$
6 \cdot 0.3 - 0.7 < k_{0} < 6 \cdot 0.3 + 0.3 \text { or } 1.1 < k_{0} < 2.1 \text { , }
$$
From this, $k_{0}=2$.
6... | 0.58 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,324 |
159. The device consists of three independently operating elements. The probabilities of faultless operation of the elements (over time $t$) are respectively: $p_{1}=0.7 ; p_{2}=0.8$; $p_{3}=0.9$. Find the probabilities that over time $t$ the elements will operate faultlessly: a) all elements; b) two elements; c) one e... | The problem is solved. The probabilities of the elements working without failure are: $p_{1}=0.7 ; p_{2}=0.8 ; p_{3}=0.9$, therefore the probabilities that the elements will fail are $q_{1}=0.3 ; q_{2}=0.2 ; q_{3}=0.1$.
50
Let's construct the generating function:
$$
\begin{aligned}
& \varphi_{3}(z)=\left(p_{1} z+q_{... | 0.504,0.398,0.092,0.006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,325 |
164. A discrete random variable $X$ is given by the distribution law:
$$
\begin{array}{ccccc}
X & 1 & 3 & 6 & 8 \\
p & 0.2 & 0.1 & 0.4 & 0.3
\end{array}
$$
Construct the distribution polygon. | Solved. We will construct a rectangular coordinate system, where along the x-axis we will plot the possible values of $x_{i}$, and along the y-axis—the corresponding probabilities $p_{i}$. We will plot the points
 has the following possible values: $x_{1}=0$ (none of the device elements failed), $x_{3}=1$ (one element failed), $x_{3}=2$ (two elements failed), and $x_{4}=3$ (three elements failed).
Since the failures of the elements are ... | \begin{pmatrix}X&0&1&2&3\\\hlinep&0.729&0.243&0.027&0.001\\\end{pmatrix} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,327 |
170. In a batch of 10 parts, there are 8 standard parts. Two parts are randomly selected. Formulate the distribution law of the number of standard parts among the selected ones *). | S o l u t i o n. The random variable $X$ - the number of standard parts among the selected parts - has the following possible values: $x_{1}=0 ; x_{2}=1 ; x_{3}=2$. We will find the probabilities of the possible values of $X$ using the formula (see problem 17, chapter $1, \S 1$ )
$$
P(X=k)=C_{n}^{k} \cdot C_{N-n}^{m-k... | \begin{pmatrix}X&0&1&2\\p&1/45&16/45&28/45\end{pmatrix} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,328 |
172. After the student answers the questions in the examination ticket, the examiner asks the student additional questions. The teacher stops asking additional questions as soon as the student shows ignorance of the given question. The probability that the student will answer any additional question is 0.9. Required: a... | Solution. a) The discrete random variable $X$ - the number of additional questions asked - has the following possible values: $x_{1}=1, x_{2}=2, x_{3}=3, \ldots, x_{k}=k, \ldots$ Let's find the probabilities of these possible values.
The variable $X$ will take the possible value $x_{i}=1$ (the examiner will ask only o... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,329 |
174. Two guns take turns firing at a target until the first hit by one of the guns. The probability of hitting the target with the first gun is 0.3, and with the second gun is 0.7. The first gun starts firing. Formulate the laws of distribution of the discrete random variables $X$ and $Y$ - the number of shells expende... | Solution. Let events $A_{i}$ and $B_{i}$ be the hits on the target by the first and second gun, respectively, at the $i$-th shot; $\bar{A}_{i}$ and $\overline{B_{i}}$ - misses.
We will find the distribution law of the random variable $X$ - the number of shells expended by the first gun. The first gun will expend one s... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,330 |
176. The textbook is published in a run of 100,000 copies. The probability that a textbook is improperly bound is 0.0001. Find the probability that the run contains exactly five defective books. | Solution. According to the condition, $n=100000, p=0.0001, k=5$. The events consisting of books being incorrectly bound are independent, the number $n$ is large, and the probability $p$ is small, so we will use the Poisson distribution
$$
P_{n}(k)=\lambda^{k} \mathrm{e}^{-\lambda} / k!
$$
Find $\lambda$:
$$
\lambda=... | 0.0375 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,331 |
179. A factory sent 500 items to a base. The probability of an item being damaged in transit is 0.002. Find the probabilities that during transit the number of damaged items will be: a) exactly three; b) less than three; c) more than three; d) at least one. | Solution. The number $n=500$ is large, the probability $p=0.002$ is small, and the events under consideration (damage to items) are independent, so the Poisson formula applies:
$$
P_{n}(k)=\lambda k \cdot e^{-\lambda} / k!
$$
a) Let's find $\lambda$:
$$
\lambda=n p=500 \cdot 0.002=1
$$
Let's find the probability th... | 0.0613,0.9197,0.019,0.632 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,332 |
181. a) The device consists of a large number of independently operating elements with the same (very small) probability of failure of each element over time $T$. Find the average number of elements that fail over time $T$, if the probability that at least one element will fail during this time is 0.98. | Solution. From the condition of the problem (since the number of elements is large, the elements work independently, and the probability of failure of each element is small), it follows that the number of failures is distributed according to the Poisson law, and it is required to find the parameter $\lambda$ (the avera... | 3.9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,333 |
182. Prove that the sum of the probabilities of the number of occurrences of an event in independent trials, calculated according to the Poisson distribution, is equal to one. It is assumed that the trials are conducted an infinite number of times. | Solution. By the Poisson law
$$
P_{n}(k)=\lambda k \mathrm{e}^{-\lambda} / k!
$$
We use the Maclaurin series expansion of the function ${ }^{x}$:
$$
\mathrm{e}^{x}=1+x / 1!+x^{2} / 2!+\ldots
$$
It is known that this series converges for any value of $x$, so by setting $x=\lambda$, we get
$$
\mathrm{e}^{\lambda}=1+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,334 |
183. The probability of winning with one lottery ticket is $p=0.01$. How many tickets need to be bought to win at least one of them with a probability $P$, not less than 0.95? | Solution. The probability of winning is small, and the number of tickets that need to be purchased is obviously large, so the number of winning tickets is approximately Poisson distributed.
It is clear that the events "none of the purchased tickets are winning" and "at least one ticket is winning" are opposite. Theref... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,335 |
184. Show that the Poisson formula, which defines the probability of $k$ events occurring over a time duration $t$
$$
P_{t}(k)=\frac{(\lambda t)^{k} \cdot \mathrm{e}^{-\lambda t}}{k!}
$$
can be considered a mathematical model of the simplest event flow; in other words, show that the Poisson formula reflects all the p... | Solution. From formula (*), it is clear that the probability of $k$ events occurring over a time duration $t$, given a certain intensity $\lambda$, is a function only of $k$ and $t$, which reflects the stationarity property of the simplest flow.
Formula (*) does not use information about the occurrence of events befor... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 32,336 |
185. The average number of taxi orders received by a dispatch center per minute is three. Find the probability that in 2 minutes: a) four calls will be received; b) fewer than four calls will be received; c) no fewer than four calls will be received. | Solution. According to the condition, $\lambda=3, t=2, k=4$. We will use the Poisson formula
$$
P_{t}(k)=\frac{(\lambda t)^{k} \cdot \mathrm{e}^{-\lambda t}}{k!}
$$
a) The required probability that four calls will be received in two minutes is
$$
P_{2}(4)=\frac{6^{4} \cdot \mathrm{e}^{-6}}{4!}=\frac{1296 \cdot 0.002... | 0.135,0.1525,0.8475 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,337 |
189. Find the mathematical expectation of the random variable $Z$, if the mathematical expectations of $X$ and $Y$ are known:
a) $Z=X+2 Y, M(X)=5, M(Y)=3 ;$ b) $Z=3 X+4 Y$, $M(X)=2, M(Y)=6$ | Solution. a) Using the properties of mathematical expectation (the expectation of a sum is equal to the sum of the expectations of the terms; a constant factor can be factored out of the expectation), we get
$$
\begin{aligned}
M(Z)=M(X+2 Y)= & M(X)+M(2 Y)=M(X)+2 M(Y)= \\
& =5+2 \cdot 3=11 .
\end{aligned}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,339 |
192. Given the list of possible values of a discrete random variable $X: x_{1}=-1, x_{2}=0, x_{3}=1$, and the known mathematical expectations of this variable and its square: $M(X)=0.1, M\left(X^{2}\right)=0.9$. Find the probabilities
$p_{1}, p_{2}, p_{3}$, corresponding to the possible values $x_{1}$, $x_{2}, x_{3}$. | Solution. Using the fact that the sum of the probabilities of all possible values of $X$ is one, and also taking into account that $M(X)=0.1, M\left(X^{2}\right)=0.9$, we can form the following system of three linear equations with respect to the unknown probabilities:
$$
\begin{gathered}
p_{1}+p_{2}+p_{3}=1,(-1) p_{1... | p_{1}=0.4,p_{2}=0.1,p_{3}=0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,340 |
196. Find the mathematical expectation of the discrete random variable $X$ - the number of throws of five dice in each of which one point appears on two dice, if the total number of throws is twenty. | Solution. We will use the formula
$$
M(X)=n P,
$$
where $n$ is the total number of trials (throws of five dice); $X$ is the number of occurrences of the event of interest (one point appears on two of the five dice) in $n$ trials ($P$ is the probability of the event of interest occurring in one trial.
According to th... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,341 |
197. The device consists of $n$ elements. The probability of failure of any element during the experiment is $p$. Find the mathematical expectation of the number of such experiments, in each of which exactly $m$ elements will fail, if a total of $\boldsymbol{N}$ experiments are conducted. It is assumed that the experim... | Solution. Let $X$ be the number of trials in which exactly $m$ elements will fail. Since the trials are independent and the probability of the event of interest (exactly $\boldsymbol{m}$ elements will fail in one trial) is the same in these trials, the formula
$$
M(X)=N P
$$
can be applied, where $N$ is the total num... | M(X)=NC_{n}^{}p^{}q^{n-} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,342 |
199. $n$ dice are thrown. Find the expected value of the sum of the number of points that will fall on all faces. | Solution. Let $X$ be the sum of the number of points that will fall on all faces, and $X_{i}(i=1,2, \ldots, n)$ be the number of points that fall on the $i$-th die face. Then, obviously,
$$
X=X_{1}+X_{2}+\ldots+X_{n}
$$
Therefore,
$$
\begin{aligned}
& M(X)=M\left(X_{1}+X_{2}+\ldots+X_{n}\right)= \\
& =M\left(X_{1}\r... | M(X)=(7/2)n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,343 |
202. Events $A_{1}, A_{2}, \ldots, A_{n}$ are mutually exclusive and form a complete group; the probabilities of these events occurring are respectively $p_{1}, p_{2}, \ldots, p_{n}$. If in the end of the trial, event $A_{i}(i=1,2, \ldots, n)$ occurs, then the discrete random variable $X$ takes the possible value $x_{i... | Solution. The possible values of the quantity $X$ according to the condition are equal to the probabilities $p_{i}$ of events $A_{i}$; the probability of a possible value $p_{i}$ is obviously also equal to $p_{i}$. Thus, $\boldsymbol{X}$ has the following distribution:
$$
\begin{array}{ccccc}
X & p_{1} & p_{2} & \ldot... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,344 |
203. Prove that the mathematical expectation of a discrete random variable is contained between its smallest and largest possible values. | Solution. Let $X$ be a discrete random variable defined by the distribution law:
$$
\begin{array}{ccc}
X & x_{1} & x_{2} \ldots x_{n} \\
p & p_{i} & p_{2} \ldots p_{n}
\end{array}
$$
Denote the smallest and largest possible values of $X$ by $m$ and $M$, respectively. Then
$$
\begin{gathered}
M(X)=x_{1} p_{1}+x_{2} p... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,345 |
204. A discrete random variable $X$ takes $k$ positive values $x_{1}, x_{2}, \ldots, x_{k}$ with probabilities equal to $p_{1}, p_{2}, \ldots, p_{k}$, respectively. Assuming that the possible values are written in increasing order, prove that
$$
\lim _{n \rightarrow \infty} \frac{M\left(X^{n+1}\right)}{M\left(X^{n}\ri... | Solution. Taking into account that
$$
P\left(X^{n+1}=x_{l}^{n+1}\right)=P\left(X=x_{i}\right)=p_{i} \text { and } P\left(X^{n}=x_{i}^{n}\right)=p_{i}
$$
we obtain
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{M\left(X^{n+1}\right)}{M\left(X^{n}\right)}=\lim _{n \rightarrow \infty} \frac{x_{1}^{n+1} p_{1}+\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,346 |
205. Prove that if random variables $X_{1}, X_{9}$, $\ldots, X_{n}$ are independent, positive, and identically distributed, then
$$
M\left[\frac{X_{1}}{X_{1}+X_{2}+\ldots+X_{n}}\right]=\frac{1}{n}
$$ | Solution. Let us introduce the random variables
$$
\begin{gathered}
Y_{1}=\frac{X_{1}}{X_{1}+X_{2}+\ldots+X_{n}}, Y_{2}=\frac{X_{2}}{X_{1}+X_{2}+\ldots+X_{n}}, \ldots, Y_{n}= \\
=\frac{X_{n}}{X_{1}+X_{2}+\ldots+X_{n}} .
\end{gathered}
$$
Note that the denominators of these fractions cannot be zero, since the values $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,347 |
207. Find the mathematical expectation of a discrete random variable $X$ distributed according to the Poisson law:
$$
\begin{array}{cccccc}
X & 0 & 1 & 2 & k & k \\
p & \mathrm{e}^{-\lambda} & \frac{\lambda \mathrm{e}^{-\lambda}}{1!} & \frac{\lambda^{2} \mathrm{e}^{-\lambda}}{2!} \cdots \frac{\lambda^{k} \mathrm{e}^{-... | Solution. By the definition of the expectation for the case when the number of possible values of $\boldsymbol{X}$ is a countable set,
$$
M(X)=\sum_{k=0}^{\infty} k \cdot \frac{\lambda k_{\mathrm{e}}-\lambda}{k!}
$$
Considering that the first term of the sum is zero when $k=0$, we take the smallest value of $k$ to be... | M(X)=\lambda | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,348 |
208. Random variables $X$ and $Y$ are independent. Find the variance of the random variable $Z=3X+2Y$, given that $D(X)=5, D(Y)=6$. | Solution. Since the variables $X$ and $Y$ are independent, the variables $3X$ and $2Y$ are also independent. Using the properties of variance (the variance of the sum of independent random variables is equal to the sum of the variances of the addends; a constant factor can be factored out of the variance sign, squaring... | 69 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,349 |
210. Find the variance and standard deviation of the discrete random variable $X$, given by the distribution law:
$$
\begin{array}{ccccc}
X & -5 & 2 & 3 & 4 \\
p & 0.4 & 0.3 & 0.1 & 0.2
\end{array}
$$ | Solution. The variance can be calculated from its definition, but we will use the formula
$$
D(X)=M\left(X^{2}\right)-[M(X)]^{2}
$$
which leads to the goal faster.
Let's find the mathematical expectation of $X$:
$$
M(X)=-5 \cdot 0.4+2 \cdot 0.3+3 \cdot 0.1+4 \cdot 0.2=-0.3
$$
Write the distribution law of $X^{2}$:... | 3.9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,350 |
212. A discrete random variable $X$ has only two possible values $x_{1}$ and $x_{2}$, and they are equally probable. Prove that the variance of the variable $X$ is equal to the square of half the difference of the possible values:
$$
\dot{D}(X)=\left[\frac{x_{2}-x_{1}}{2}\right]^{2}
$$ | Solution. Let's find the expected value of $X$, considering that the probabilities of possible values $x_{1}$ and $x_{2}$ are equal to each other and, therefore, each of them is equal to $1 / 2$:
$$
M(X)=x_{1} \cdot(1 / 2)+x_{2} \cdot(1 / 2)=\left(x_{1}+x_{2}\right) / 2 .
$$
Let's find the expected value of $X^{2}$:
... | [\frac{x_{2}-x_{1}}{2}]^{2} | Algebra | proof | Yes | Yes | olympiads | false | 32,351 |
213. Find the variance of the discrete random variable $X$ - the number of occurrences of event $A$ in five independent trials, if the probability of event $A$ occurring in each trial is 0.2. | Solution. The variance of the number of occurrences of an event in independent trials (with the same probability of the event occurring in each trial) is equal to the product of the number of trials by the probabilities of the event occurring and not occurring:
$$
D(X)=n p q .
$$
By the condition, $n=5 ; p=0.2 ; q=1-... | 0.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,352 |
215. Find the variance of a discrete random variable $X$ - the number of occurrences of event $A$ in two independent trials, if the probabilities of the event occurring in these trials are the same and it is known that $M(X)=1.2$. | Solution. First method. The possible values of the quantity $X$ are: $x_{1}=0$ (the event did not occur), $x_{2}=1$ (the event occurred once), and $x_{3}=2$ (the event occurred twice).
We will find the probabilities of the possible values using the Bernoulli formula: $P_{2}(0)=q^{2} ; \quad P_{2}(1)=C_{2}^{1} p q=2 p ... | 0.48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,353 |
218. A discrete random variable $X$ has only two possible values: $x_{1}$ and $x_{2}$, with $x_{2}>x_{1}$. The probability that $X$ will take the value $x_{1}$ is 0.6. Find the distribution law of the variable $X$, if the expected value and variance are known: $M(X)=1.4$; $D(X)=0.24$. | Solution. The sum of the probabilities of all possible values of a discrete random variable is equal to one, so the probability that $X$ will take the value $x_{2}$ is $1-0.6=0.4$.
Let's write the distribution law of $X$:
$$
\begin{array}{lll}
X & x_{1} & x_{2} \\
p & 0.6 & 0.4
\end{array}
$$
To find $x_{1}$ and $x_... | \begin{pmatrix}X&1&2\\p&0.6&0.4\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,354 |
221. $n$ dice are thrown. Find the variance of the sum of the number of points that can appear on all the fallen faces. | Solution. Let $X$ be the discrete random variable representing the sum of the number of points that will fall on all faces, and $X_{i}(i=1,2, \ldots, n)$ be the number of points that fall on the $i-\hat{\text { th }}$ die. Then
$$
X=X_{1}+X_{2}+\ldots+X_{n}
$$
Obviously, all variables $X_{i}$ have the same distributi... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,355 | |
224. Prove the inequality $M\left[X-\left(x_{i}+x_{k}\right) / 2\right]^{2} \geqslant D(X)$, where $x_{i}$ and $x_{k}$ are any two possible values of the random variable $X$. | Solution. 1) Suppose that $\left(x_{i}+x_{k}\right) / 2=M(X)$. Then
$$
M\left[X-\frac{x_{i}-x_{k}}{2}\right]^{2}=D(X)
$$
2) Suppose that $\left(x_{i}+x_{k}\right) / 2 \neq M(X)$. We will prove that in this case
$$
M\left[X-\frac{x_{i}+x_{k}}{2}\right]^{2}>D(X)
$$
Transform the left side of the inequality using the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,356 |
225. Prove that if a random variable $X$ has the smallest and largest possible values, which are $a$ and $b$ respectively, then the variance of this random variable does not exceed the square of half the difference between these values:
$$
D(X) \leqslant[(b-a) / 2]^{2}
$$ | Solution. We will use the inequality (see problem 224)
$$
D(X)<M[X-(a+b) / 2]^{2} .
$$
We will now prove that
$$
M[X-(a+b) / 2]^{2} \leq[(b-a) / 2]^{2} .
$$
(From this and (") follows the validity of the inequality to be proved.) For this purpose, we transform the expectation:
$$
\begin{aligned}
& M[(b-a) / 2]^{2}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,357 |
226. Prove that if $X$ and $Y$ are independent random variables, then
$$
D(X Y)=D(X) \cdot D(Y)+n^{2} D(X)+m^{2} D(Y)
$$
where $m=M(X)$ and $n=M(Y)$. | Solution. By the formula for calculating variance
$$
D(X Y)=M\left[(X Y)^{2}\right]-[M(X Y)]^{2}
$$
Considering that $X$ and $Y$ are independent variables, and therefore $X^{2}$ and $Y^{2}$ are also independent, and that the expected value of the product of independent random variables is equal to the product of thei... | D(XY)=D(X)D(Y)+n^{2}D(X)+^{2}D(Y) | Algebra | proof | Yes | Yes | olympiads | false | 32,358 |
230. A discrete random variable $X$ is given by the distribution law:
$$
\begin{array}{lccc}
X & 1 & 2 & 4 \\
p & 0.1 & 0.3 & 0.6
\end{array}
$$
Find the central moments of the first, second, third, and fourth orders. | Solution. The first-order central moment is zero: $\boldsymbol{\mu}_{i}=\mathbf{0}$.
For the calculation of central moments, it is convenient to use formulas expressing central moments in terms of initial moments, so we will first find the initial moments:
$$
\begin{aligned}
& v_{1}=M(X)=1 \cdot 0.1+2 \cdot 0.3+4 \cd... | \mu_1=0,\mu_2=1.29,\mu_3=-0.888,\mu_4=2.7777 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,361 |
232. Prove that the central moment of the second order (variance) $\mu_{2}=M[X-M(X)]^{2}$ is less than the ordinary moment of the second order $\mu_{2}^{\prime}=M[X-C]^{2}$ for any $C \neq M(X)$. | Solution. For the sake of simplicity, let's denote $M(X)=\boldsymbol{m}$. Add and subtract $m$ under the sign of the expected value:
$$
\begin{aligned}
& \mu_{2}^{\prime}=M[X-C]^{2}=M[(X-m)+(m-C)]^{2}= \\
& =M\left[(X-m)^{2}+2(m-C)(X-m)+(m-C)^{2}\right]
\end{aligned}
$$
The expected value of a sum is equal to the sum... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,362 |
233. Prove that the central moment of the third order is related to the initial moments by the equality
$$
\mu_{3}=v_{3}-3 v_{1} v_{2}+2 v_{1}^{3}
$$ | Solution. By the definition of the central moment,
$$
\mu_{3}=M[X-M(X)]^{3}
$$
Using the properties of mathematical expectation and considering that $M(X)$ is a constant value, we get
$$
\begin{gathered}
\mu_{8}=M\left[X^{3}-3 X^{2} \cdot M(X)+3 X \cdot M^{2}(X)-M^{3}(X)\right]= \\
=M\left(X^{3}\right)-3 M(X) \cdot ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,363 |
235. Let $X=X_{1}+X_{2}$, where $X_{1}$ and $X_{2}$ are independent random variables with central moments of the third order, equal to $\mu_{3}^{1}$ and $\mu_{3}^{2}$, respectively. Prove that $\mu_{3}=\mu_{3}^{1}+\mu_{3}^{2}$, where $\mu_{3}$ is the central moment of the third order of the variable $X$. | Solution. For the sake of simplicity, let's introduce the following notations for the expectations: $M\left(X_{1}\right)=a_{1}, M\left(X_{2}\right)=a_{2}$. Then
$$
M(X)=M\left(X_{1}+X_{2}\right)=M\left(X_{1}\right)+M\left(X_{2}\right)=a_{1}+a_{2} .
$$
By definition, the third central moment is
$$
\begin{gathered}
\m... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,364 |
241. The device consists of 10 independently operating elements: The probability of failure of each element over time $T$ is 0.05. Using Chebyshev's inequality, estimate the probability that the absolute value of the difference between the number of failed elements and the average (expected) number of failures over tim... | Solution. a) Let $X$ be a discrete random variable - the number of failed elements over time $T$. Then
$$
\begin{aligned}
& M(X)=n p=10 \cdot 0.05=0.5 \\
& D(X)=n p q=10 \cdot 0.05 \cdot 0.95=0.475 .
\end{aligned}
$$
Using Chebyshev's inequality:
$$
P(|X-M(X)|<\varepsilon) \geqslant 1-D(X) / \varepsilon^{2}
$$
Subs... | 0.880.12 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 32,365 |
243. The probability of event $A$ occurring in each trial is 1/2. Using Chebyshev's inequality,
estimate the probability that the number $X$ of occurrences of event $A$ is within the range from 40 to 60, if 100 independent trials are conducted. | Solution. Let's find the mathematical expectation and variance of the discrete random variable $X$ - the number of occurrences of event $A$ in 100 independent trials:
$$
M(X)=n p=100 \cdot 1 / 2=50 ; D(X)=n p q=100 \cdot 1 / 2 \cdot 1 / 2=25
$$
Let's find the maximum difference between the given number of occurrences... | 0.75 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 32,366 |
245. A discrete random variable $X$ is given by the distribution law
$$
\begin{array}{lll}
X & 0.3 & 0.6 \\
p & 0.2 & 0.8
\end{array}
$$
Using Chebyshev's inequality, estimate the probability that $|X-M(X)|<0.2$. | Solution. Let's find the mathematical expectation and variance of the quantity $X$:
$$
\begin{gathered}
M(X)=0.3 \cdot 0.2 + 0.6 \cdot 0.8 = 0.54; \\
D(X)=M\left(X^{2}\right)-[M(X)]^{2}= \\
=\left(0.3^{2} \cdot 0.2 + 0.6^{2} \cdot 0.8\right) - 0.54^{2} = 0.0144.
\end{gathered}
$$
We will use Chebyshev's inequality in... | 0.64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 32,367 |
247. A sequence of independent random variables $X_{1}, X_{2}, \ldots, X_{n}, \ldots$ is given by the distribution law
| $X_{n}$ | $-n \alpha$ | 0 | $n \alpha$ |
| :---: | :---: | :---: | :---: |
| $p$ | $1 /\left(2 n^{2}\right)$ | $1-1 / n^{2}$ | $1 /\left(2 n^{2}\right)$ |
Is Chebyshev's theorem applicable to the g... | Solution. For the Chebyshev's theorem to be applicable to a sequence of random variables, it is sufficient that these variables are pairwise independent, have finite mathematical expectations, and uniformly bounded variances.
Since the random variables are independent, they are a fortiori pairwise independent, i.e., t... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,368 |
249. A sequence of independent random variables \(X_{1}, X_{2}, \ldots, X_{n}, \ldots\) is given by the distribution law
\[
\begin{array}{lcc}
X_{n} & n+1 & -n \\
p & n /(2 n+1) & (n+1) /(2 n+1)
\end{array}
\]
a) Verify that the requirement of the Chebyshev theorem for uniform boundedness of variances is not satisfie... | Solution. Since the random variables $X_{n}$ are independent, they are also pairwise independent, i.e., the first requirement of Chebyshev's theorem is satisfied.
It is easy to find that $M\left(X_{n}\right)=0$, i.e., the requirement of the finiteness of mathematical expectations is satisfied.
It remains to check the... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,369 |
252. The random variable $X$ is given by the distribution function
$$
F(x)=\left\{\begin{array}{llr}
0 & \text { for } & x \leqslant-1 \\
(3 / 4) x+3 / 4 & \text { for } & -11 / 3
\end{array}\right.
$$
Find the probability that in the result of the trial, the variable $X$ will take a value in the interval $(0,1 / 3)$... | Solution. The probability that $X$ will take a value within the interval $(a, b)$ is equal to the increment of the distribution function over this interval: $P(a<X<b)=F(b)-F(a)$. Setting $a=0$, $b=1 / 3$, we get
$$
\begin{gathered}
P(0<X<1 / 3)=F(1 / 3)-F(0)= \\
=[(3 / 4) x+3 / 4]_{x=1 / 3}-[(3 / 4) x+3 / 4]_{x=0}=1 /... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,370 |
258. The random variable $X$ is defined on the entire $O x$ axis by the distribution function $F(x)=1 / 2+(1 / \pi) \operatorname{arctg}(x / 2)$. Find the possible value $x_{1}$, satisfying the condition:
with probability $1 / 4$ the random variable $X$ will take a value greater than $x_{i}$ in the experiment. | Solution. Events $Xx_{1}$ are opposite, so $P\left(Xx_{1}\right)=1$. Therefore, $P\left(Xx_{1}\right)=1-1 / 4=3 / 4$. Since $P\left(X=x_{1}\right)=0$, then
$$
P\left(X<x_{1}\right)=P\left(X=x_{1}\right)+P\left(X<x_{1}\right)=P\left(X<x_{1}\right)=3 / 4
$$
By the definition of the distribution function,
$$
P\left(X<x... | x_{1}=2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,371 |
260. A discrete random variable $X$ is given by the distribution law
| $X$ | 2 | 4 | 7 |
| :---: | :---: | :---: | :---: |
| $p$ | 0.5 | 0.2 | 0.3 |
Find the distribution function $F(x)$ and draw its graph. | Solution. 1. If $x<2$, then $\boldsymbol{F}(\boldsymbol{x})=0$. Indeed, the value of $X$ does not take values less than the number 2. Therefore, for $x \leqslant 2$ the function $F(x)=P(X<x)=0$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,372 |
271. The density function of a continuous random variable $X$ is given on the entire $O x$ axis by the equation $f(x)=$ $=4 C /\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)$. Find the constant parameter $C$. | Solution. The density function $f(x)$ must satisfy the condition $\int_{-\infty}^{\infty} f(x) \mathrm{d} x=1$. We require that this condition be satisfied for the given function:
$$
4 C \int_{-\infty}^{\infty} \frac{\mathrm{d} x}{\mathrm{e}^{x}+\mathrm{e}^{-x}}=1
$$
From this,
$$
C=1 /\left(4 \int_{-\infty}^{\infty... | \frac{1}{2\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,376 |
277. The random variable $X$ in the interval (-c, c) is given by the probability density function $f(x)=1 /\left(\pi \sqrt{c^{2}-x^{2}}\right)$; outside this interval, $f(x)=0$. Find the mathematical expectation of the variable $X$. | Solution. We use the formula $M(X)=\int_{a} x f(x) \mathrm{d} x$. Substituting $a=-c, b=c, f(x)=1 /\left(\pi \sqrt{c^{2}-x^{2}}\right)$, we get
$$
M(X)=\frac{1}{\pi} \int_{-c}^{c} \frac{x d x}{\sqrt{c^{2}-x^{2}}}
$$
Considering that the integrand is an odd function and the limits of integration are symmetric with res... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,377 |
280. Find the mathematical expectation of the random variable $X$, given by the distribution function
$$
F(x)=\left\{\begin{array}{ccr}
0 & \text { for } & x \leqslant 0 \\
x / 4 & \text { for } & 0 < x \leqslant 4 \\
1 & \text { for } & x > 4
\end{array}\right.
$$ | Solution. Find the density function of the random variable $\boldsymbol{X}$:
$$
f(x)=F^{\prime}(x)=\left\{\begin{array}{ccc}
0 & \text { for } & x4 .
\end{array}\right.
$$
Find the required expected value:
$$
M(X)=\int_{0}^{4} x f \cdot(x) \mathrm{d} x=\int_{0}^{4} x \cdot(1 / 4) \mathrm{d} x=2
$$ | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,378 |
285. The random variable $X$ is given by the probability density function $f(x)=2 \cos 2x$ in the interval $(0, \pi / 4)$; outside this interval, $f(x)=0$. Find: a) the mode; b) the median of $X$. | Solution. a) It is easy to verify that the function $f(x)=2 \cos 2 x$ in the open interval $(0, \pi / 4)$ does not have a maximum, therefore $X$ does not have a mode.
b) Let's find the median $M_{e}(X)=m_{e}$, based on the definition of the median!
$P\left(Xm_{e}\right)$, or equivalently, $P\left(X<m_{e}\right)=1 / 2... | m_{e}=\pi/12 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,379 |
286. The random variable $X$ in the interval $(2,4)$ is given by the probability density function $f(x)=-(3 / 4) x^{2}+(9 / 2) x-6$; outside this interval, $f(x)=0$. Find the mode, the expected value, and the median of the variable $X$. | Solution. Represent the density function as $f(x)=$ $=-(3 / 4)(x-3)^{2}+3 / 4$. From this, it is clear that the density function reaches its maximum at $x=3$; therefore, $M_{0}(X)=3$. (Of course, the maximum could have been found using methods of differential calculus.)
The distribution curve is symmetric with respect... | M_{0}(X)=3,M(X)=3,M_{e}(X)=3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,380 |
290. Prove that the mathematical expectation of a continuous random variable is between its smallest and largest possible values. | Solution. Let $X$ be a continuous random variable defined by the probability density function $f(x)$ on the interval $[a, b]$; outside this interval, $f(x)=0$. Then $a<x<b$. Considering that $f(x) \geqslant 0$, we get $a f(x)<x f(x)<b f(x)$. Integrating this double inequality from $a$ to $b$:
$$
a \int_{a}^{b} f(x) \m... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,381 |
295. A random variable $X$ in the interval $(0, \pi)$ is given by the probability density function $f(x)=(1 / 2) \sin x$; outside this interval, $f(x)=0$. Find the variance of $X$. | Solution. We will find the variance using the formula
$$
D(X)=\int_{a}^{b} x^{2} f(x) d x-[M(X)]^{2}
$$
Substituting here $M(X)=\pi / 2$ (the distribution curve is symmetric about the line $x=\pi / 2$), $a=0, b=x, f(x)=(1 / 2) \sin x$, we get
$$
D(X)=\frac{1}{2} \int_{0}^{\pi} x^{2} \sin x d x-\left[\frac{\pi}{2}\ri... | \frac{\pi^{2}-8}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,383 |
297. Find the variance of the random variable $X$, given by the distribution function
$$
F(x)=\left\{\begin{array}{ccc}
0 & \text { for } & x \leqslant-2 \\
x / 4+1 / 2 & \text { for } & -2 < x \leqslant 2 \\
1 & \text { for } & x > 2
\end{array}\right.
$$ | Solution. Let's find the density function:
$$
f(x)=F^{\prime}(x)=\left\{\begin{array}{ccc}
0 & \text { for } & x2
\end{array}\right.
$$
Let's find the expected value
$$
M(X)=\int_{-2}^{2} x f(x) \mathrm{d} x=\int_{-2}^{2} x \cdot \frac{1}{4} \mathrm{~d} x=0
$$
(the integrand function is odd, and the limits of integ... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,384 |
301. A random variable $X$ is given by the probability density function $f(x)=x^{n} \mathrm{e}^{-x} / n!$ for $x \geqslant 0 ; f(x)=0$ for $x<0$. Find: a) the expected value; b) the variance of $X$. | Solution. a) Let's find the mathematical expectation:
$$
M(X)=\int_{0}^{\infty} x f(x) \mathrm{d} x=\frac{1}{n!} \int_{0}^{\infty} x \cdot x^{n} \mathrm{e}^{-x} \mathrm{~d} x=\frac{1}{n!} \int_{0}^{\infty} x^{n+1} \mathrm{e}^{-x} d x
$$
We will use the so-called gamma function, which is defined by the equation
$$
\G... | D(X)=n+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,385 |
303. Prove that for any continuous random variable, the central moment of the first order is equal to zero. | Solution. By the definition of the first-order central moment,
$$
\mu_{i}=\int_{-\infty}^{\infty}[x-M(X)] f(x) \mathrm{d} x=\int_{-\infty}^{\infty} x f(x) \mathrm{d} x-M(X) \int_{-\infty}^{\infty} f(x) \mathrm{d} x
$$
Considering that
$$
\int_{-\infty}^{\infty} x f(x) \mathrm{d} x=M(X) \quad \mathrm{h} \quad \int_{-... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,386 |
304. Prove that the second-order raw moment
$$
\mu_{2}^{\prime}=\int_{-\infty}^{\infty}(x-c)^{2} f(x) \mathrm{d} x
$$
has the smallest value if $c=M(X)$. | Solution. Transform $\mu_{z}^{\prime}$ as follows:
$$
\begin{gathered}
\mu_{2}^{\prime}=\int_{-\infty}^{\infty}(x-c)^{2} f(x) \mathrm{d} x=\int_{-\infty}^{\infty}[(x-M(X))+ \\
+(M(X)-c)]^{2} f(x) \mathrm{d} x=\int_{-\infty}^{\infty}[x-M(X)]^{2} f(x) \mathrm{d} x+ \\
+2[M(X)-c] \int_{-\infty}^{\infty}[x-M(X)] f(x) \mat... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,387 |
305. A random variable $X$ is given by the density function $f(x)=0.5 x$ in the interval (0,2); outside this interval $f(x)=0$. Find the raw and central moments of the first, second, third, and fourth orders. | Solved and done. By the formula
$$
v_{k}=\int_{0}^{2} x^{k} f(x) \mathrm{d} x
$$
we find the initial moments:
$$
\begin{array}{ll}
v_{1}=\int_{0}^{2} x \cdot(0.5 x) \mathrm{d} x=\frac{4}{3} ; \quad v_{8}=\int_{0}^{2} x^{2} \cdot(0.5 x) \mathrm{d} x=2 \\
v_{3}=\int_{0}^{2} x^{3} \cdot(0.5 x) \mathrm{d} x=3.2 ; & v_{4... | \mu_{2}=\frac{2}{9},\mu_{3}=-\frac{8}{135},\mu_{4}=\frac{16}{135} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,388 |
308. The scale division value of the amperemeter is 0.1 A. The readings of the amperemeter are rounded to the nearest whole division. Find the probability that an error exceeding $0.02 \mathrm{~A}$ will be made during the reading. | Solution. The rounding error of the measurement can be considered as a random variable $X$, which is uniformly distributed in the interval between two adjacent integer divisions. The density of the uniform distribution $f(x)=1 /(b-a)$, where $(b-a)$ is the length of the interval in which the possible values of $X$ are ... | 0.6 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,389 |
313. Find the mathematical expectation of a random variable $X$, uniformly distributed in the interval (a,b). | Solution. The density function of a uniform distribution is symmetric about the line $x=(a+b) / 2$, so $M(X)=(a+b) / 2$.
Thus, the expected value of a random variable uniformly distributed over the interval ( $a, b$ ) is the half-sum of the endpoints of this interval. Of course, the same result can be obtained using t... | (+b)/2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,390 |
315. Find the variance and standard deviation of a random variable $X$ that is uniformly distributed in the interval $(a, b)$. | It is solved. We use the formula
$$
D(X)=\int_{a}^{b} x^{2} f(x) \mathrm{d} x-[M(X)]^{2}
$$
Substituting $f(x)=1 /(b-a), M(X)=(a+b) / 2$ (see problem 313) and performing elementary calculations, we obtain the desired variance
$$
D(X)=(b-a)^{2} / 12
$$
The mean square deviation of the random variable $X$ is equal to... | D(X)=\frac{(b-)^2}{12},\sigma(X)=\frac{b-}{2\sqrt{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,391 |
318. The diameter of a circle $x$ is measured approximately, with $a \leqslant x \leqslant b$. Considering the diameter as a random variable $X$, uniformly distributed in the interval $(a, b)$, find the mathematical expectation and variance of the area of the circle. | Solution. 1. Find the expected value of the area of the circle - a random variable $Y=\varphi(K)=\pi X^{2} / 4$ - using the formula
$$
M[\varphi(X)]=\int_{a}^{b} \varphi(x) f(x) d x
$$
Substituting $\varphi(x)=\pi x^{2} / 4, f(x)=1 /(b-a)$ and performing the integration, we get
$$
M\left[\pi X^{2} / 4\right]=\pi\lef... | \pi(b^{2}++^{2})/12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,392 |
328. The expected value and standard deviation of a normally distributed random variable $X$ are 10 and 2, respectively. Find the probability that in the experiment, $X$ will take a value in the interval $(12,14)$. | Solution. We will use the formula
$$
P(\alpha<X<\beta)=\Phi\left(\frac{\beta-a}{\sigma}\right)-\Phi\left(\frac{\alpha-a}{\sigma}\right)
$$
Substituting $\alpha=12, \beta=14, a=10$ and $\sigma=2$, we get $P(12<X<14)=$ $=\Phi(2)-\Phi(1)$. From the table in Appendix 2, we find: $\Phi(2)=0.4772$, $\Phi(1)=0.3413$. The de... | 0.1359 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,394 |
331. The measurement of the shaft diameter is being performed without systematic (one-sided) errors. Random measurement errors $X$ follow a normal distribution with a root mean square deviation $\sigma=10$ mm. Find the probability that the measurement will be performed with an error not exceeding $15 \mathrm{mm}$ in ab... | Solution. The mathematical expectation of random errors is zero, so the formula $P(|X|<\delta)=2 \Phi(\delta / \sigma)$ can be applied. Setting $\delta=15, \sigma=10$, we find $P(|X|<15)=2 \Phi(1.5)$. From the table in Appendix 2, we find: $\Phi(1.5)=0.4332$. The required probability
$$
P(|X|<15)=2 \cdot 0.4332=0.8664... | 0.8664 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,395 |
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