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Example 1. A die is rolled. Let's denote the events: $A$ - "rolling a six", $B$ - "rolling a three", $C$ - "rolling an even number", $D$ - "rolling a number that is a multiple of three". What are the relationships between these events? | Solution. If six points fall, then an even number of points also falls, i.e., event $A$ implies event $C: A \subset C$. Reasoning similarly, we get $A \subset D, B \subset D, A+B=D, C \cdot D=A$. | A\subsetC,A\subsetD,B\subsetD,A+B=D,C\cdotD=A | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,930 |
Example 2. Experiment - tossing a die. Events: $A_{k}(k=1,2,3,4,5,6)$ - "$k$ points fall", $A$ - "an even number of points falls", $B$ - "an odd number of points falls", $C$ - "a number of points that is a multiple of three falls", $D$ - "a number of points greater than three falls". Express events $A, B, C$ and $D$ in... | The solution is as follows. Event $A$ occurs if and only if $A_{2}$, or $A_{4}$, or $A_{6}$ occurs. This means that $A=A_{2}+A_{4}+A_{6}$. Reasoning in a similar manner, we get: $\quad B=A_{1}+A_{3}+A_{5}, \quad C=A_{3}+A_{6}$, $D=A_{4}+A_{5}+A_{6}$. | A=A_{2}+A_{4}+A_{6},\quadB=A_{1}+A_{3}+A_{5},\quadC=A_{3}+A_{6},\quadD=A_{4}+A_{5}+A_{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,931 |
Example 3. Let $A, B, C$ be arbitrary events. What do the following events mean: $\bar{A} B C, \bar{A} \bar{B} \bar{C}, \bar{A}+\bar{B}+\bar{C}, A \bar{B} \bar{C}+\bar{A} B \bar{C}+\bar{A} \bar{B} C$, $\bar{A} \bar{B} \bar{C}+\bar{A} \bar{B} C+\bar{A} B \bar{C}+A \bar{B} \bar{C}?$ | Solution. According to the definition, $\bar{A} B C$ is the product of three events $\bar{A}, B, C$ that occur simultaneously, where $\bar{A}$ is the event opposite to event $A$: Therefore, $\bar{A} B C$ means that event $A$ did not occur, while events $B$ and $C$ did occur. Reasoning similarly, we conclude that: $\ove... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,932 |
Example 4. The experiment consists of a shooter firing 3 shots at a target. Event $A_{k}$ - "hit the target on the $k$-th shot $(k=1,2,3)$". Express the following events in terms of $A_{1}, A_{2}, A_{3}$: $A$ - "at least one hit"; $B$ - "three hits"; $C$ - "three misses"; $D$ - "at least one miss"; $E$ - "not less than... | Solution. Event $A$ occurs if and only if $A_{1}$, or $A_{2}$, or $A_{3}$ occurs. This means that $A=A_{1}+A_{2}+A_{3}$. Three hits will occur if and only if a hit occurs with each shot, i.e., events $A_{1}, A_{2}, A_{3}$ all occur together: $B=A_{1} \cdot A_{2} \cdot A_{3}$. Three misses will occur if and only if a mi... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,933 |
Example 5. Experiment - extracting a part from a box containing items of three grades. Event notations: $A$ - "a first-grade part is extracted", $B$ - "a second-grade part is extracted", $C$ - "a third-grade part is extracted". What do the following events represent: $A+B ; \overline{A+C} ; A C, A B+C?$ | Solution. $A+B$ is the event that occurs when at least one of the events $A$ and $B$ happens. Therefore, $A+B$ in this case is a part of the first or second grade. Since $A+C$ is a part of the first or third grade, the opposite of this event $\overline{A+C}$ is a part of the second grade. $A C$ is an impossible event, ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,934 |
Example 6. Prove that $\overline{A+B}=\bar{A} \cdot \bar{B}$.
Solution. To prove this equality, it is sufficient to show that $\overline{A+B} \subset \bar{A} \cdot \bar{B}$ and $\bar{A} \cdot \bar{B} \subset \overline{A+B}$. If the event $\overline{A+B}$ occurs, this means that the event opposite to $A+B$ has occurred... | Solution. Taking into account the equalities (1.6.1), (1.6.4), (1.6.5), $(1.6 .10),(1.6 .11)$, we obtain
$$
\begin{aligned}
& (A+C)(B+C)=A(B+C)+C(B+C)=A B+A C+C B+C C= \\
& =A B+(A+B) C+C=A B+(A+B) C+C U=A B+(A+B+U) C= \\
& =A B+U C=A B+C
\end{aligned}
$$
(Here $U$ is a certain event). | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,935 |
Example 8. Simplify the expression $(A+B)(A+\bar{B})$. | Solution. Denoting the certain event by $U$, the impossible event by $V$, and using formulas (1.6.1), (1.6.4), (1.6.5) - (1.6.8) and $(1.6 .11)$, we get
$$
\begin{aligned}
& (A+B)(A+\bar{B})=A(A+\bar{B})+B(A+\bar{B})=A A+A \bar{B}+B A+B \bar{B}= \\
& =A+A(B+\bar{B})+V=A+A U+V=A+A+V=A+V=A
\end{aligned}
$$
Thus, $(A+B)... | A | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,936 |
Example 9. Simplify the expression $(\bar{A}+B)(\bar{A}+\bar{B})$. | Solution. Using formulas (1.6.1), (1.6.4), (1.6.5) - (1.6.8), (1.6.11), we find that
$$
\begin{aligned}
& (\bar{A}+B)(\bar{A}+\bar{B})=\bar{A}(\bar{A}+\bar{B})+B(\bar{A}+\bar{B})=\overline{A A}+\bar{A} \bar{B}+B \bar{A}+B \bar{B}= \\
& =\overline{A A}+\bar{A}(\bar{B}+B)+B \bar{B}=\bar{A}+\bar{A} U+V=\bar{A}+\bar{A}+V=... | \bar{A} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,937 |
Example 1. Prove that the sum of the probabilities of opposite events is equal to one. | Solution. Since $A+\dot{\bar{A}}=\Omega, A$ and $\bar{A}$ are mutually exclusive events, then by Axiom 3
$$
P(A+\bar{A})=P(A)+P(\bar{A})
$$
Since $\Omega$ is a certain event, then
$$
P(\Omega)=1 \text { (by Axiom } 2 \text { ). }
$$
Furthermore,
$$
P(A+\bar{A})=P(\Omega)
$$
Taking into account the two previous eq... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,938 |
Example 2. Prove that the probability of an impossible event is zero. | Solution. In equality (1.7.8), let $A=\Omega, \bar{A}=\varnothing$, then
$$
P(\Omega)+P(\varnothing)=1
$$
Since by axiom 2 $P(\Omega)=1$, then
$$
P(\varnothing)=0
$$ | P(\varnothing)=0 | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 32,939 |
Example 3. Prove that for pairwise disjoint events $A_{1}, A_{2}, \ldots, A_{n}$, the following equality holds
$$
P\left(A_{1}+A_{2}+\ldots+A_{n}\right)=P\left(A_{1}\right)+P\left(A_{2}\right)+\ldots+P\left(A_{n}\right)
$$ | Solution. For $n=2$, this equality holds (by Axiom 3).
Assume that it is true for $n=k-1$; we will prove that it will also hold for $n=k$. Indeed,
\[
\begin{aligned}
& P\left(A_{1}+A_{2}+\ldots+A_{k-1}+A_{k}\right)=P\left(\left(A_{1}+A_{2}+\ldots+A_{k-1}\right)+A_{k}\right)= \\
& =P\left(A_{1}+A_{2}+\ldots+A_{k-1}\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,940 |
Example 4. Prove that for any events $A$ and $B$, the addition theorem of probabilities holds:
$$
P(A+B)=P(A)+P(B)-P(A B)
$$ | Solution. Let us represent the events $A+B$ and $B$ as the corresponding sums of mutually exclusive events $A+B=A+B \bar{A}, B=B \bar{A}+B A$ (see formula (1.6.12)). Applying Axiom 3, we get
$$
P(A+B)=P(A)+P(B \bar{A}), P(B)=P(B \bar{A})+P(A B)
$$
By determining $P(B \bar{A})$ from the second equality and substitutin... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,941 |
Example 6. Prove that if event $A$ implies event $B$ ($A \subset B$), then
$$
P(A) \leq P(B)
$$ | Solution. In this case $B=A+\bar{A} B$ (see formula (1.6.13)), therefore
$$
P(B)=P(A)+P(\bar{A} B)
$$
Since $P(\bar{A} B) \geq 0$, then $P(B) \geq P(A)$, or $P(A) \leq P(B)$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,942 |
Example 7. Prove that the probability of any event $A$ satisfies the inequalities
$$
0 \leq P(A) \leq 1
$$ | Solution. Since $\varnothing \subset A \subset \Omega, P(\varnothing)=0, P(\Omega)=1$, it follows from (1.7.13) that
$$
P(\varnothing) \leq P(A) \leq P(\Omega)
$$
i.e., $0 \leq P(A) \leq 1$. Thus, the probability of any event is expressed by a non-negative number not exceeding unity; in other words, all values of the... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 32,943 |
Example 8. Two dice are rolled. What is the probability that the sum of the points on both dice will not exceed 5? | Solution. Let $n_{1}$ points fall on the first die, $n_{2}$ - on the second. The space of elementary events is the set of pairs $(n_{1}, n_{2})$:
$$
\Omega=\left\{\left(n_{1}, n_{2}\right): n_{1}, n_{2}=1,2,3,4,5,6\right\}
$$
Event $A$ is of the form
$$
A=\left\{\left(n_{1}, n_{2}\right): n_{1}, n_{2}=1,2,3,4 ; n_{1... | \frac{5}{18} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,944 |
Example 1. A die is rolled. What is the probability that an even number of points will come up? | Solution. Let's introduce the notation: $A$ - an even number of points has fallen; $B_{k}$ - $k$ points have fallen ( $k=1,2,3,4,5,6$ ). Event $A$ means that at least one of the events: $B_{2}, B_{4}, B_{6}$ has occurred, i.e., $A=B_{2}+B_{4}+B_{6}$. Since the events $B_{2}, B_{4}, B_{6}$ are mutually exclusive, we can... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,945 |
Example 2. In an urn, there are 40 balls: 15 blue, 5 green, and 20 white. What is the probability that a colored ball will be drawn from the urn? | Solution. Extracting a colored ball means the appearance of either a blue or a green ball. The probability of extracting a blue ball (event $A$): $P(A)=15 / 40=3 / 8$. The probability of extracting a green ball (event $B$): $P(B)=5 / 40=1 / 8$. Since events $A$ and $B$ are mutually exclusive, we obtain by formula (1.8.... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,946 |
Example 3. Two dice are rolled. Find the probability of event $A$ - "the sum of the points rolled does not exceed four". | Solution. Event $A$ is the sum of three mutually exclusive events $B_{2}, B_{3}, B_{4}$, which means the sum of the points is 2, 3, or 4, respectively. Since
$$
P\left(B_{2}\right)=\frac{1}{36}, \quad P\left(B_{3}\right)=\frac{2}{36}, \quad P\left(B_{4}\right)=\frac{3}{36}
$$
by the theorem of addition of probabiliti... | \frac{1}{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,947 |
Example 4. A sportsman shoots at a target divided into 3 sectors. The probability of hitting the first sector is 0.4, the second - 0.3. What is the probability of hitting either the first or the second sector? | Solution. Events $A$ - "hitting the first sector" and $B$ - "hitting the second sector" are mutually exclusive (hitting one sector excludes hitting the other), so the addition rule for probabilities of mutually exclusive events applies. According to this rule, the required probability is:
$$
P(A+B)=P(A)+P(B)=0.4+0.3=0... | 0.97 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,948 |
Example 6. A fair coin is tossed three times. What is the probability that heads will come up exactly twice? | Solution. Let's introduce the notation: $A_{k}$ - "the appearance of a digit on the $k$-th coin toss $(k=1,2,3)$ ", $A$ - "the appearance of two digits in three tosses", then
$$
A=A_{1} A_{2} \bar{A}_{3}+A_{1} \bar{A}_{2} A_{3}+\bar{A}_{1} A_{2} A_{3}
$$
Since the terms on the right side of this equation are pairwise... | \frac{3}{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,949 |
Example 7. From the first machine, 200 parts were sent to assembly, of which 190 are standard; from the second - 300, of which 280 are standard. Find the probability of event $A$, which consists in a randomly taken part being standard, and the conditional probabilities of it relative to events $B$ and $\bar{B}$, if eve... | Solution. The probability of event $A$ is the ratio of the number of all standard parts to the total number of parts manufactured on both machines
$$
P(A)=\frac{190+280}{200+300}=\frac{470}{500}=0.94
$$
The conditional probability of event $A$ given event $B$ (the probability that a randomly taken part is standard, g... | 0.94,0.95,\frac{14}{15}\approx0.93 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,950 |
Example 8. An urn contains 8 red and 6 blue balls. Three balls are drawn sequentially without replacement from the urn. Find the probability that all 3 balls are blue. | Solution. Let's introduce the notations: $A_{1}$ - "the first ball is blue", $A_{2}$ - "the second ball is blue", $A_{3}$ - "the third ball is blue", $A$ - "all 3 balls are blue", then $A=A_{1} A_{2} A_{3}$. We will use formula (1.8.11), which for $n=3$ takes the form
$$
P\left(A_{1} A_{2} A_{3}\right)=P\left(A_{1}\ri... | \frac{5}{91} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,951 |
Example 12. In each of the three boxes, there are 30 parts. In the first box, there are 27, in the second 28, and in the third 25 standard parts. One part is randomly taken from each box. What is the probability that all three parts taken will be standard. | Solution. The probability that a standard part is drawn from the first box (event $A$), $P(A)=27 / 30=9 / 10$. The probability that a standard part is drawn from the second box (event $B$) $P(B)=28 / 30=14 / 15$. The probability that a standard part is drawn from the third box (event $C$) $P(C)=25 / 30=5 / 6$. Since ev... | 0.7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,953 |
Example 13. There are two urns with balls of three colors. The first contains 2 blue, 3 red, and 5 green balls, while the second contains 4 blue, 2 red, and 4 green balls. One ball is drawn from each urn and their colors are compared. Find the probability that the colors of the drawn balls are the same (event $A$). | Solution. Let the event of drawing a blue ball from the first urn be denoted by $B_{1}$, a red ball by $C_{1}$, and a green ball by $D_{1}$. Similar events for the second urn will be denoted by $B_{2}, C_{2}, D_{2}$ respectively. Event $A$ occurs in the case of $B_{1} B_{2}, C_{1} C_{2}$, or $D_{1} D_{2}$, i.e., $A=B_{... | 0.34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,954 |
Example 14. A worker services four identical machines. The probability that any machine will require the worker's attention during an hour is 0.6. Assuming that machine malfunctions are independent, find the probability that during an hour: a) all four machines; b) no machine; c) at least one machine. | Solution. Let $A_{1}, A_{2}, A_{3}, A_{4}$ be the events that the first, second, third, and fourth machines, respectively, will require the worker's attention within an hour. According to the multiplication theorem of probabilities of independent events, the probability that all machines will require the worker's atten... | 0.9744 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,955 |
Example 17. In an urn, there are 6 blue and 4 red balls. Two balls are drawn in succession. What is the probability that both balls are blue? | Solution. Let event $A$ be "drawing a blue ball on the first draw", and event $B$ be "drawing a blue ball on the second draw". We need to find the probability of event $A B$. Since
$$
\begin{gathered}
P(A)=\frac{6}{10}=\frac{3}{5} \\
P(B / A)=\frac{6-1}{10-1}=\frac{5}{9}
\end{gathered}
$$
$$
P(A B)=P(A) \cdot P(B / A... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,956 |
Example 18. In a workshop, two motors are working independently of each other. The probability that the first motor will not require the master's attention during an hour is 0.85, and for the second motor, this probability is 0.8. Find the probability that during the hour neither of the motors will require the master's... | Solution. We introduce notations for events: $A$ - "the first motor will not require the master's attention during an hour", $B$ - "the second motor will not require attention during an hour". Let's find the probability of the event $A B$. Since $A$ and $B$ are independent events, then $\cdot$
$$
P(A B)=P(A) P(B)=0.85... | 0.68 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,957 |
Example 19. On 30 identical tokens, 30 two-digit numbers from 1 to 30 are written. The tokens are placed in a bag and thoroughly mixed. What is the probability of drawing a token with a number that is a multiple of 2 or 3? | Solution. Let's denote the events: $A$ - "a token with an even number is drawn", $B$ - "a token with a number divisible by 3 is drawn"; $A B$ - "a token with an even number divisible by 3 is drawn". We need to find the probability of the event $A+B$. Since $A$ and $B$ are compatible events, then
$$
P(A+B)=P(A)+P(B)-P(... | \frac{2}{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 32,958 |
Example 20. From an urn containing 3 blue and 2 red balls, balls are drawn sequentially according to a random selection scheme without replacement. Find the probability $P_{k}$ that a red ball will appear for the first time on the $k$-th trial ( $k=1,2,3,4$ ). | Solution. We introduce the following notations for events: $A_{k}$ - "a red ball appears at the $k$-th trial", $B_{k}$ - "the first red ball appears at the $k$-th trial" ($k=1,2,3,4$). The events $B_{k}$ can be expressed through $A_{i}$ and $\bar{A}_{i}$: $B_{1}=A_{1}, B_{2}=\bar{A}_{1} A_{2}, B_{3}=\bar{A}_{1} \bar{A}... | p_{1}=0.4,p_{2}=0.3,p_{3}=0.2,p_{4}=0.1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,959 |
Example 22. How many times do you need to roll two dice so that the probability of rolling at least one double six is greater than $1 / 2$? (This problem was first posed by the French mathematician and writer de Mere ( $1610-1684$ ), hence the problem is named after him). | Solution. Let the event $A_{i}$ be "rolling two sixes on the $i$-th throw". Since any of the six faces of the first die can match any of the six faces of the second die, there are $6 \cdot 6=36$ equally likely and mutually exclusive events. Only one of these - rolling a six on both the first and the second die - is fav... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,960 |
Example 23. The probability that an event will occur at least once in three independent trials is 0.973. Find the probability of the event occurring in one trial (it is assumed that the probability of the event occurring is the same in all trials).
| Solution. Since the events under consideration are independent in aggregate, formula (1.8.17) is applicable, i.e.
$$
P(A)=1-q^{n}
$$
By the condition
$$
P(A)=0.973, n=3
$$
Therefore,
$$
0.973=1-q^{3}
$$
or
$$
\begin{aligned}
& q^{3}=1-0.973=0.027 \\
& q=\sqrt[3]{0.027}=0.3
\end{aligned}
$$
The required probabil... | 0.7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,961 |
Example 24. In the box, there are 15 balls, of which 5 are blue and 10 are red. Two balls are drawn sequentially from the box; the first ball is not returned to the box. Find the probability that the first ball drawn will be blue and the second will be red. | Solution. Let $A$ be the event "the first ball is blue", and $B$ the event "the second ball is red". From the condition, we have
$$
P(A)=\frac{5}{15}=\frac{1}{3}, \quad P(B / A)=\frac{10}{14}=\frac{5}{7}
$$
According to the first of the formulas (1.8.8), we get
$$
P(A B)=P(A) P(B / A)=\frac{5}{15} \cdot \frac{10}{14... | \frac{5}{21} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,962 |
Example 25. The word papakha is made up of letters from a cut-out alphabet. The cards with the letters are thoroughly shuffled. Four cards are drawn in sequence and laid out in a row. What is the probability of getting the word nana this way? | Solution. Let $A, B, C, D$ denote the events of drawing the first, second, third, and fourth letter of the word "papa" from a set of 6 letters: $a, a, a, n, n, x$. We need to find the probabilities of the events: $A, B / A, C / A B, D / A B C$.
$$
P(A)=\frac{2}{6}=\frac{1}{3}
$$
$$
\begin{aligned}
& P(B / A)=\frac{3}... | \frac{1}{30} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,963 |
Example 1. At a factory manufacturing bolts, the first machine produces $30 \%$, the second - $25 \%$, the third - $45 \%$ of all products. The defect rate in their products is $2 \%$, $1 \%$, $3 \%$ respectively. Find the probability that a randomly selected bolt is defective. | Solution. Let $A$ be the event that a randomly selected bolt is defective, and let $H_{1}, H_{2}, H_{3}$ be the events that this bolt is produced by the first, second, and third machines, respectively. From the problem statement, we have
$$
\begin{aligned}
& P\left(H_{1}\right)=0.30, P\left(H_{2}\right)=0.25, P\left(H... | 0.022 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,964 |
Example 2. In five boxes, there are balls of the same size and weight. In two boxes - 6 blue and 4 red balls each (this is box composition $H_{1}$). In two other boxes (composition $H_{2}$) - 8 blue and 2 red balls each. In one box (composition $H_{3}$) - 2 blue and 8 red balls. A box is randomly chosen and a ball is d... | Solution. Let the event "a red ball is drawn" be denoted by $A$.
From the problem statement, it follows that
$$
P\left(H_{1}\right)=\frac{2}{5}=0.4, \quad P\left(H_{2}\right)=\frac{2}{5}=0.4, \quad P\left(H_{3}\right)=\frac{1}{5}=0.2 .
$$
The probability of drawing a red ball, given that the box is from the first ba... | 0.4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,965 |
Example 3. A batch of electric bulbs is manufactured by the first factory at $20 \%$, by the second factory at $30 \%$, and by the third factory at $50 \%$. The probabilities of producing defective bulbs are respectively: $q_{1}=0.01, q_{2}=0.005, q_{3}=0.006$. Find the probability that a randomly selected bulb from th... | Solution. We introduce the following notations: $A$ - "a standard bulb is taken from the batch", $H_{1}$ - "the taken bulb is manufactured by the first factory", $H_{2}$ - "by the second factory", $H_{3}$ - "by the third factory". We find the conditional probabilities $P\left(A / H_{i}\right)(i=1,2,3)$ using the formul... | 0.9935 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,966 |
Example 4. In a group of 21 students, including 5 excellent students, 10 well-performing students, and 6 poorly performing students. On the upcoming exam, excellent students can only receive excellent grades. Well-performing students can receive good or excellent grades with equal probability. Poorly performing student... | Solution. Let's denote the hypotheses: $H_{1}$ - "an excellent student is invited", $H_{2}$ - "a good student is invited", $H_{3}$ - "a weak student is invited".
From the problem statement, it follows that
$$
\begin{aligned}
& P\left(H_{1}\right)=\frac{5}{21}, P\left(H_{2}\right)=\frac{10}{21}, P\left(H_{3}\right)=\f... | \frac{17}{21} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,967 |
Example 5. Parts from three machines arrive at the assembly. It is known that the first machine produces $0.1\%$ of defective parts, the second - $0.2\%$, the third - $0.3\%$. Find the probability of a defective part arriving at the assembly, if the first machine supplied 1000, the second - 2000, and the third - 3000 p... | Solution. We introduce the following notations: event $A$ - "a defective part has arrived"; hypothesis $H_{1}$ - "the part was manufactured on the first machine", $H_{2}$ - "on the second", $H_{3}$ - "on the third".
From the problem statement, it follows that
$$
\begin{aligned}
& P\left(H_{1}\right)=\frac{1000}{6000}... | 0.0023 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,968 |
Example 6. A worker services 3 machines on which identical parts are processed. The probability of defect for the first machine is 0.02, for the second - 0.03, for the third - 0.04. Processed parts are stored in one box. The productivity of the first machine is three times greater than that of the second, and the third... | Solution. Let's introduce the following notations: event $A$ - "a defective part is taken"; hypothesis $H_{1}$ - "the part is made on the first machine", $H_{2}$ - "the part is made on the second machine", $H_{3}$ - "the part is made on the third machine".
Let $x$ be the productivity of the second machine, then $3 x$ ... | 0.024 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,969 |
Example 7. At two automatic machines, identical parts are manufactured. It is known that the productivity of the first machine is twice that of the second, and that the probability of producing a part of the highest quality on the first machine is 0.9, and on the second - 0.81. Parts manufactured during the shift on bo... | Solution. Let $A$ be the event that a randomly selected part will turn out to be of the highest quality. Event $A$ can occur with one of the following hypotheses: $H_{1}$ - "the part is manufactured on the first machine", $H_{2}$ - "the part is manufactured on the second machine".
Since the productivity of the first m... | 0.87 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,970 |
Example 8. At a distribution base, there are electric bulbs manufactured by two factories. Among them, $60 \%$ are made by the first factory and $40 \%$ by the second. It is known that out of every 100 bulbs made by the first factory, 95 meet the standard, and out of 100 bulbs made by the second factory, 85 meet the st... | Solution. Let $A$ be the event that the randomly selected bulb is standard. We introduce two hypotheses: $H_{1}$ "the bulb is manufactured by the first factory", $H_{2}$ - "the bulb is manufactured by the second factory".
From the conditions of the problem, it follows that
$$
\begin{aligned}
& P\left(H_{1}\right)=0.6... | 0.91 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,971 |
Example 9. A radio tube may belong to one of three batches with probabilities: $p_{1}=0.2, p_{2}=0.3, p_{3}=0.5$. The probability that the tube will work for a given number of hours, for these batches, is respectively: 0.9; 0.8; 0.7. Determine the probability that the radio tube will work for the given number of hours. | Solution. Let's introduce the following notations: $A$ - the event that the lamp will work for the specified number of hours; $H_{1}, H_{2}, H_{3}$ - hypotheses that the lamp belongs to the first, second, and third batch, respectively.
According to the problem, we have:
$$
\begin{aligned}
& P\left(H_{1}\right)=0.2, \... | 0.77 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,972 |
Example 10. At an enterprise, products of a certain type are manufactured on three production lines. The first line produces $30 \%$ of the products from the total production volume, the second line - $25 \%$, and the third line produces the remaining part of the products. Each line is characterized by the following pe... | Solution. Let's introduce the following notations: $A$ - the event that a randomly selected item is defective; $H_{1}, H_{2}, H_{3}$ - hypotheses that the item was produced on the first, second, and third production lines, respectively.
First, note that 45% of the items are produced on the third line from the total pr... | 0.032 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,973 |
Example 11. In a box, there are identical items manufactured by two machines: $40 \%$ of the items are made by the first machine, the rest - by the second. Defects in the production of the first machine constitute $3 \%$, of the second $-2 \%$. Find the probability that a randomly selected item will be defective. | Solution. Let $A$ be the event that a randomly selected item is defective; and let $H_{1}, H_{2}$ be the events that the item is manufactured by the first and second machine, respectively.
From the condition, it follows that
$$
\begin{aligned}
& P\left(H_{1}\right)=0.4, \quad P\left(H_{2}\right)=0.6, \quad\left(P\lef... | 0.024 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,974 |
Example 12. There are three urns with balls. The first contains 5 blue and 3 red balls, the second - 4 blue and 4 red, and the third - 8 blue. One of the urns is randomly chosen, and a ball is randomly drawn from it. What is the probability that it will be red (event $A$). | Solution. The ball can be drawn from the first urn, or the second, or the third. Let's denote the selection of the first, second, and third urns by $H_{1}, H_{2}, H_{3}$, respectively.
Since there are equal chances of selecting any of the urns, then
$$
P\left(H_{1}\right)=P\left(H_{2}\right)=P\left(H_{3}\right)=\frac... | \frac{7}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,975 |
Example 1. In a box, there are identical items manufactured by two machines: $40 \%$ of the items are made by the first machine, the rest - by the second. Defects in the production of the first machine constitute $3 \%$, and of the second $-2 \%$. Find the probability that a randomly selected item was made by the first... | Solution. Let $A$ be the event that a randomly selected item is defective, and let $H_{1}, H_{2}$ be the events that the item was manufactured by the first and second machine, respectively.
From the condition, it follows that
$$
\begin{aligned}
& P\left(H_{1}\right)=0.4, \quad P\left(H_{2}\right)=1-P\left(H_{1}\right... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,976 |
Example 2. Parts manufactured by two plants are stored in a warehouse. It is known that the production volume of the first plant exceeds the production volume of the second plant by 4 times. The probability of defect at the first plant is $p_{1}=0.05$, at the second plant - $p_{2}=0.01$. A randomly selected part turned... | Solution. Let $H_{1}$ be the event that the part taken is manufactured at the first plant, $H_{2}$ - at the second plant, then
$$
P\left(H_{1}\right)=\frac{4}{5}=0.8, P\left(H_{2}\right)=\frac{1}{5}=0.2
$$
Let $A$ be the event that a randomly taken part turned out to be defective.
According to the condition
$$
P\le... | 0.952 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,977 |
Example 3. Products from three factories are delivered to a warehouse, where products from the first factory account for $20 \%$, the second - $46 \%$, and the third $34 \%$. It is known that the average percentage of non-standard items for the first factory is $3 \%$, for the second $-2 \%$, and for the third - $1 \%$... | Solution. Let $A$ be the event that a non-standard item is taken, and let $H_{1}, H_{2}, H_{3}$ be the hypotheses that the item taken was manufactured at the first, second, and third factory, respectively.
From the problem statement, it follows that
$$
\begin{aligned}
& P\left(H_{1}\right)=0.20, \quad P\left(H_{2}\ri... | 0.322 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,978 |
Example 4. Machines $a, b, c$ on a factory produce $20\%, 35\%, 45\%$ of all items respectively. In their production, the defect rate is $3\%, 2\%$, $4\%$. What is the probability that a randomly selected defective item was produced by machines $a, b, c$ respectively? | Solution. Let event $A$ consist in the fact that a randomly selected item is defective, and $H_{1}, H_{2}, H_{3}$ be events that the item is produced by machines $a, b, c$ respectively. Events $H_{1}, H_{2}, H_{3}$ form a complete system of events. The numbers 0.20; 0.35; 0.45 (20%, 35%, 45%) are the probabilities of t... | 0.1936,0.2258,0.5806 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,979 |
Example 5. A certain product is manufactured by two factories. The production volume of the second factory exceeds that of the first by 3 times. The defect rate of the first factory is $2 \%$, and that of the second is $1 \%$. Products manufactured by the factories over the same period of time were mixed and sent for s... | Solution. Let $A$ be the event that a defective item has been purchased, and let $H_{1}$ and $H_{2}$ be the events that the item was produced by the first and second factory, respectively.
Since the volume of production of the second factory is three times the volume of production of the first, then
$$
P\left(H_{1}\r... | 0.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,980 |
Example 6. In five boxes, there are balls of the same weight and size. In two boxes - 6 blue and 4 red balls each (this is box composition $H_{1}$). In two other boxes (composition $H_{2}$) - 8 blue and 2 red balls each. In one box (composition $H_{3}$) - 2 blue and 8 red balls. A box is randomly chosen and a ball is d... | Solution. Let $A$ be the event that a blue ball is drawn from the box.
From the problem statement, it follows that
$$
P\left(H_{1}\right)=\frac{2}{5}=0.4 ; \quad P\left(H_{2}\right)=\frac{2}{5}=0.4 ; \quad P\left(H_{3}\right)=\frac{1}{5}=0.2
$$
The probability of drawing a blue ball, given that the box chosen is $H_... | 0.4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,981 |
Example 7. At an enterprise, products of a certain type are manufactured on three production lines. The first line produces $30 \%$ of the products from the total production volume, the second line - $25 \%$, and the third line - the remaining part of the production. Each line is characterized by the following percenta... | Solution. Let's introduce the following notations: $A$ - the event that a randomly selected item is defective; $H_{1}, H_{2}, H_{3}$ - hypotheses that the item was manufactured on the first, second, and third lines, respectively.
According to the problem statement, we have:
$$
\begin{aligned}
& P\left(H_{1}\right)=0.... | 0.281,0.156,0.563 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,982 |
Example 8. In the first urn, there are 2 blue and 6 red balls, in the second urn, there are 4 blue and 2 red balls. Two balls are randomly transferred from the first urn to the second, after which one ball is randomly drawn from the second urn.
What is the probability that this ball is blue?
Assuming the ball drawn f... | Solution. Let's introduce the following notations: event $A$ - "the ball drawn from the second urn is blue"; hypotheses $H_{1}$ - "two blue balls were transferred from the first urn to the second", $H_{2}$ - "two balls of different colors were transferred", $H_{3}$ - "two red balls were transferred".
We will calculate... | \frac{1}{21} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,983 |
Example 3. A discrete random variable $X$ has the following distribution:
| $X$ | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.2 | 0.4 | $p_{4}$ | 0.1 |
What is the probability $p_{4}=P \quad(X=0.8) ?$
Construct the distribution polygon. | Solution. Since equality (2.2.1) must hold, i.e.,
$$
p_{1}+p_{2}+p_{3}+p_{4}+p_{5}=1
$$
then
$$
\begin{gathered}
p_{4}=1-\left(p_{1}+p_{2}+p_{3}+p_{5}\right)=1-(0.1+0.2+0.4+0.1)=1-0.8=0.2 \\
p_{4}=0.2
\end{gathered}
$$
$ and $p_{3}=P(X=5)$, given that $p_{3}$ is 4 times $p_{1}$. | Solution. Since
$$
p_{2}+p_{4}+p_{5}=0.15+0.25+0.35=0.75
$$
then based on equality (2.1.2), we conclude that
$$
p_{1}+p_{3}=1-0.75=0.25
$$
Since by the condition $p_{3}=4 p_{1}$, then
$$
p_{1}+p_{3}=p_{1}+4 p_{1}=5 p_{1} .
$$
Thus, $5 p_{1}=0.25$, from which $p_{1}=0.05$; therefore,
$$
p_{3}=4 p_{1}=4 \cdot 0.05... | p_{1}=0.05;p_{3}=0.20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,986 |
Example 5. Two fair coins are tossed, and the number of heads on both upper sides of the coins is counted. Consider the discrete random variable $X$ - the number of heads that appear on both coins. Write the distribution law of the random variable $X$. | Solution. In this experiment, there are four equally likely elementary outcomes: $(\Gamma, \Gamma),(\Gamma, L),(Ц, \Gamma),(Ц, L);$ the notation $(\Gamma, L)$ means that the coat of arms fell on the first coin, and the tail on the second; the other notations have a similar meaning. The coat of arms can fall 1 time, 2 t... | \begin{pmatrix}X&0&1&2\\\hlineP&0.25&0.50&0.25\\\end{pmatrix} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,987 |
Example 6. Two dice are rolled, and the number of points that fall on both top faces is counted. Find the distribution law of the discrete random variable $X$ - the sum of the points that fall on two dice.
---
The provided translation maintains the original text's formatting and structure. | Solution. In this trial, there are 36 equally possible elementary outcomes (see Table 1.1). The random variable $X$ can take integer values from 2 to 12, where the values 2 and 12 occur once, 3 and 11 - twice, 4 and 10 - three times, 5 and 9 - four times, 6 and 8 - five times, and the value 7 - six times.
Let's calcul... | \begin{pmatrix}X&2&3&4&5&6&7&8&9&10&11&12\\\hlineP&\frac{1}{36}&\frac{1}{18}&\frac{1}{12}&\frac{1}{9}&\frac{5}{36} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,988 |
Example 8. The probability of producing a non-standard item with a certain technological process is 0.06. The inspector takes an item from the batch and immediately checks its quality. If it turns out to be non-standard, further tests are stopped, and the batch is held. If the item is standard, the inspector takes the ... | Solution. The discrete random variable $X$ can take five values: $x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=5$. It will take the value $x_{1}=1$, i.e., only one item will be checked and the batch will be detained, if the first item checked by the inspector turns out to be non-standard. The probability of such an outcom... | \begin{pmatrix}X&1&2&3&4&5\\\hlineP&0.06&0.056&0.053&0.050&0.781\\\end{pmatrix} | Other | math-word-problem | Yes | Yes | olympiads | false | 32,989 |
Example 9. A series of independent trials is conducted, in each of which event $A$ occurs with the same probability $p$. The trials continue until the first occurrence of event $A$. The random variable $X$ - the number of trials conducted - is considered. Formulate the distribution law for it. | Solution. The specified random variable $X$ can take values $x_{1}=1, x_{2}=2, x_{3}=3, \ldots$. The event $X=n(n=1,2,3, \ldots)$ means that the event $A$ does not occur in the first $n-1$ trials, but occurs in the $n$-th trial. The probability of such an outcome is:
$$
\underbrace{q \cdot q \cdot \ldots \cdot q}_{n-1... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,990 |
Example 3. Is the function
$$
F(x)=\frac{1}{1+x^{2}}(-\infty<x<+\infty) ?
$$
a distribution function of a random variable? | Solution. The given function is not a distribution function of a random variable, as it decreases in the interval $(0,+\infty)$; inequality (2.2.4) is not satisfied in this interval. The graph of the function is shown in Fig. 2.6.
Note that all values of the given function belong to the interval $(0,1]$, i.e., the fun... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,992 |
Example 4. Random variables $X_{1}$ and $X_{2}$ have distribution functions $F_{1}(x)$ and $F_{2}(x)$, respectively. Prove that the function
$$
F(x)=C_{1} F_{1}(x)+C_{2} F_{2}(x)
$$
is the distribution function of some random variable $X$, where $C_{1}$ and $C_{2}$ are non-negative numbers whose sum is one. | Solution. Since $F_{1}(x)$ and $F_{2}(x)$ are distribution functions, they satisfy conditions (2.2.3) - (2.2.5), (2.2.7).
Taking these conditions into account, we obtain:
$$
\begin{gathered}
0 \leq F(x)=C_{1} F_{1}(x)+C_{2} F_{2}(x) \leq C_{1}+C_{2}=1 ; 0 \leq F(x) \leq 1 \\
F\left(x_{1}\right)=C_{1} F_{1}\left(x_{1}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,993 |
Example 5. The distribution law of a discrete random variable is given by the following table:
| $X$ | 0 | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.2 | 0.4 | 0.3 | 0.1 |
Find the distribution function of this random variable. | Solution. To construct the distribution function $F(x)$ of a discrete random variable $X$, we use formula (2.2.11).
1. For $x \leq 0 \quad F(x)=\sum_{x_{k}3 \quad F(x)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=$
$$
=0.2+0.4+0.3+0.1=1
$$
The graph of the function $F(x)$ is shown in Fig. 2.7. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,994 |
Example 6. In a batch of 10 parts, there are 8 standard parts. Two parts are randomly taken from this batch. Find the distribution function of the discrete random variable equal to the number of standard parts in the sample. | Solution. First, we find
the distribution law of the given random variable $X$. This variable can take three values: $x_{1}=0, x_{2}=1, x_{3}=2$. We will calculate the probabilities of these values:
$$
\begin{gathered}
p_{1}=P(X=0)=\frac{C_{8}^{0} \cdot C_{2}^{2}}{C_{10}^{2}}=\frac{1}{45}, p_{2}=P(X=1)=\frac{C_{8}^{1... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,995 |
Example 7. The random variable $X$ is given by the distribution function
$$
F(x)=\left\{\begin{array}{lc}
0 & \text { if } x \leq 0 \\
x / 2 & \text { if } 0 < x \leq 2 \\
1 & \text { if } x > 2
\end{array}\right.
$$
Find the probability that in the experiment the random variable $X$ will take a value in the interval... | Solution. For this interval, $F(x)=x / 2$. According to formula (2.2.2), we get:
$$
P(1<X<2)=F(2)-F(1)=(2 / 2)-(1 / 2)=0.5
$$ | 0.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,996 |
Example 8. The random variable $X$ is given by the distribution function
$$
F(x)=\left\{\begin{array}{lc}
0 & \text { if } x \leq 0 \\
x / 3 & \text { if } 0 < x \leq 3 \\
1 & \text { if } x > 3
\end{array}\right.
$$
Find the probability that in the result of the trial, the variable $X$ will take a value from the int... | Solution. By formula (2.2.2), we find:
$$
P(2<x \leq 3)=F(3)-F(2)=(3 / 3)-(2 / 3)=1-2 / 3=1 / 3 .
$$ | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,997 |
Example 10. Two shooters make one shot each at one target. The probability of hitting the target for the first shooter with one shot $p_{1}=0.5$, for the second - $p_{2}=0.4$. The discrete random variable $X$ is the number of hits on the target. Find the distribution function of this random variable. Find the probabili... | Solution. First, let's find the distribution law of the given discrete random variable $X$. This variable can take three values: $x_{1}=0, x_{2}=1, x_{3}=2$.
We introduce the notation: event $A_{1}$ - "the first shooter hits", event $A_{2}$ - "the second shooter hits", then $\bar{A}_{1}$ and $\bar{A}_{2}$ are their re... | 0.7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,998 |
Example 1. The density function of a random variable $X$ is given by
106
$$
p(x)=\frac{c}{1+x^{2}}
$$
Find the value of the parameter $c$. | Solution. The density function must satisfy condition $(2.3 .6)$, i.e., the equality
$$
\int_{-\infty}^{+\infty} \frac{c}{1+x^{2}} d x=c \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}=1
$$
must hold, from which
$$
c=1 / \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}
$$
The indefinite integral is a standard one:
$$
\in... | \frac{1}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,999 |
Example 2. The probability density of a random variable $X$ is given by the function
$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x / 2 & \text { if } 0 < x \leq 2\end{cases}
$$
Find the probability that in the experiment, the variable $X$ will take a value from the interval $(1,2)$. | Solution. The required probability will be found using formula (2.3.3):
$$
P(1<X<2)=\int_{1}^{2} \frac{x}{2} d x=\left.\frac{x^{2}}{4}\right|_{1} ^{2}=\frac{2^{2}}{4}-\frac{1^{2}}{4}=1-\frac{1}{4}=\frac{3}{4}=0.75
$$ | 0.75 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,000 |
Example 3. The distribution function of a random variable $X$ is given by
$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ \frac{x^{2}}{1+x^{2}} & \text { if } x>0\end{cases}
$$
Find its probability density function. | Solution. The probability density function $p(x)$ and the distribution function $F(x)$ are related by equation (2.3.5).
According to equality (2.3.5), we find:
$$
\begin{gathered}
p(x)=F^{\prime}(x)=\left(\frac{x^{2}}{1+x^{2}}\right)^{\prime}=\frac{2 x\left(1+x^{2}\right)-2 x \cdot x^{2}}{\left(1+x^{2}\right)^{2}}=\f... | p(x)=\begin{cases}0&\text{for}x\leq0\\\frac{2x}{(1+x^{2})^{2}}&\text{for}x>0\end{cases} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,001 |
## Example 5. Given the function
$$
f(x)= \begin{cases}0 & \text { if } x \leq 0 \\ c e^{-\alpha x} & \text { if } x>0 \quad(\alpha>0)\end{cases}
$$
For what value of the constant $c$ is the function $f(x)$ a probability density function of some random variable $X$? | Solution. First of all, it must be $c \geq 0$. To determine the value of $c$, we will use condition (2.3.6):
$$
\int_{-\infty}^{+\infty} f(x) d x=1, c \int_{0}^{+\infty} e^{-\alpha x} d x=1,-\left.\frac{c}{\alpha} e^{-\alpha x}\right|_{0} ^{+\infty}=1, \frac{c}{\alpha}=1, c=\alpha
$$
Therefore, the probability densit... | \alpha | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,002 |
Example 6. Find the distribution function $F(x)$ of the random variable $X$, the probability density of which is defined by the function
$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \text{ or } x>2 \\ x & \text { if } 0<x \leq 1 \\ 2-x & \text { if } 1<x \leq 2\end{cases}
$$ | Solution. To find the distribution function $F(x)$, we will use formula (2.3.2).
For $x \leq 0 \quad$ we get $F(x)=\int_{-\infty}^{0} 0 d x=0$.
For $0 < x \leq 2$ we get $F(x)=\int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{0} p(t) d t+\int_{0}^{x} p(t) d t=$
=\begin{cases}0&\text{if}x\leq0\\x^{2}/2&\text{if}0<x\leq2\\1&\text{if}x>2\end{cases} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,003 |
Example 8. The probability density function of a random variable $X$ is given by
$$
p(x)=a x^{2} e^{-k x} \quad(k>0, \quad 0 \leq x<+\infty)
$$
Find the value of the coefficient $a$. Find the distribution function $F(x)$ of the variable $X$. | Solution. The value of the coefficient $a$ is determined from the equalities:
$$
\begin{aligned}
& \int_{0}^{+\infty} a x^{2} e^{-k x} d x=1 \\
& a=1 / \int_{0}^{+\infty} x^{2} e^{-k x} d x
\end{aligned}
$$
By integrating by parts twice, we find:
$$
\begin{aligned}
& \int_{0}^{+\infty} x^{2} e^{-k x} d x=-\left.\fra... | =\frac{k^3}{2},\quadF(x)=1-\frac{k^2x^2+2kx+2}{2}e^{-kx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,005 |
Example 9. The function $f(x)=a e^{-|x|}$ is given. For what value of $a$ can it be considered as the probability density function of some random variable $X$? | Solution. If $f(x)$ is a probability density, then condition (2.3.6) must be satisfied. Therefore,
$$
\begin{aligned}
& \int_{-\infty}^{+\infty} a e^{-|x|} d x=a \int_{-\infty}^{0} e^{x} d x+a \int_{0}^{+\infty} e^{-x} d x=\left.a e^{x}\right|_{-\infty} ^{0}-\left.a e^{-x}\right|_{0} ^{+\infty}= \\
& =a(1-0)-a(0-1)=2 ... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,006 |
Example 10. The random variable $X$ is given by the distribution function
$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ \frac{(1-\cos x)}{2} & \text { if } 0 < x \leq \pi\end{cases}
$$
Find the probability density function of the random variable $X$. Calculate the probability that the random variable $X$ will ta... | Solution. The probability density $p(\bar{x})$ and the distribution function $F(x)$ of a random variable $X$ are related by (2.3.5), i.e., $F^{\prime}(x)=p(x)$. Therefore,
$$
p(x)=0 \text { for } x=0 \text { and } x>\pi, p(x)=\left(\frac{1-\cos x}{2}\right)^{\prime}=\frac{\sin x}{2}
$$
in the interval $(0, \pi)$. Usi... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,007 |
Example 12. The random variable $X$ is given by the distribution function
$$
F(x)= \begin{cases}0 & \text { if } x \leq-a \\ \frac{(a+x)^{2}}{2 a^{2}} & \text { if }-a<x \leq a \\ 1 & \text { if } x>a .\end{cases}
$$
Find the probability density function of the random variable $X$. | Solution. Using equality (2.3.5), we find the function $p(x)$.
Since
$$
\begin{aligned}
& \left(\frac{(a+x)^{2}}{2 a^{2}}\right)^{\prime}=\frac{1}{a}+\frac{x}{a^{2}}=\frac{1}{a}\left(1+\frac{x}{a}\right) \\
& \left(1-\frac{(a-x)^{2}}{2 a^{2}}\right)^{\prime}=\frac{1}{a}\left(1-\frac{x}{a}\right)
\end{aligned}
$$
the... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,008 |
Example 2. The distribution law of a discrete random variable is given by the table
| $X$ | -4 | -2 | 0 | 2 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.2 | 0.15 | 0.25 | 0.3 |
Write the distribution laws of the random variables $3X$, $X / 2$. Find the mathematical expectations of the rando... | Solution. Let's write down the distribution laws of the random variables $3X$ and $X/2$ using tables:
| $3X$ | -12 | -6 | 0 | 6 | 12 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.2 | 0.15 | 0.25 | 0.3 |
| $X/2$ | -2 | -1 | 0 | 1 | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1... | M(X)=0.9,M(3X)=2.7,M(X/2)=0.45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,009 |
Example 3. The mathematical expectations of two random variables $X$ and $Y$ are known: $M(X)=3, M(Y)=2$.
Find the mathematical expectations of the sum and difference of these variables. | Solution. Based on formulas (2.4.12) and (2.4.14), we conclude that
$$
\begin{aligned}
& M(X+Y)=M(X)+M(Y)=3+2=5 \\
& M(X-Y)=M(X)-M(Y)=3-2=1
\end{aligned}
$$ | M(X+Y)=5,\M(X-Y)=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,010 |
Example 4. The mathematical expectations of two independent random variables $X$ and $Y$ are known: $M(X)=4, M(Y)=5$.
Find the mathematical expectation of their product. | Solution. Applying formula (2.4.15), we find
$$
M(X \cdot Y)=M(X) \cdot M(Y)=4 \cdot 5=20
$$ | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,011 |
Example 5. Find the mathematical expectation of the random variable $Y=2X+7$, given that $M(X)=4$. | Solution. Using formulas (2.4.10), (2.4.11), (2.4.12), we find
$$
\begin{aligned}
& M(Y)=M(2 X+7)=M(2 X)+M(7)= \\
& =2 M(X)+7=2 \cdot 4+7=15
\end{aligned}
$$ | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,012 |
Example 6. A die is rolled. Find the mathematical expectation of the discrete random variable $X$, equal to the number of points rolled. | Solution. This random variable can take six values: $x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=5, x_{6}=6$; the probability of each of them is the same, equal to $1 / 6$. The distribution law of the random variable $X$ can be given by the formulas
$$
\begin{gathered}
P\left(X=x_{k}\right)=1 / 6, k=1,2,3,4,5,6 \\
\sum_... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,013 |
Example 7. Two dice are rolled. The discrete random variable $X$ is the sum of the points that appear on both dice. Find the mathematical expectation of this random variable. | Solution. This random variable takes all integer values from 2 to 12. The distribution law can be given by the following table:
| $X$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | $\frac{1}{36}$ | $\frac{2}{36}... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,014 |
Example 8. Independent trials are conducted, in each of which event $A$ occurs with probability $p$. The trials continue until the first occurrence of event $A$. The random variable $X$ is the number of trials conducted. Find the mathematical expectation of the random variable $X$. | Solution. Possible values of this random variable: $x_{n}=n, \quad n=1,2,3, \ldots$. The event $X=n$ means that the event $A$ does not occur in the first $n-1$ trials, but occurs in the $n-$th trial. The probability of such an outcome is
$$
P(X=n)=\underbrace{q q \ldots q}_{n-1 \text { times }} \cdot p=p q^{n-1}, \qua... | \frac{1}{p} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,015 |
Example 10. The probability density function of a random variable $X$ is given by
$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 3 x^{2} / 8 & \text { if } \quad 0 < x < 2\end{cases}
$$
Find the mathematical expectation of the random variable $X$. | Solution. Using formula (2.4.7), we find
$$
M(X)=\int_{0}^{2} x \cdot 3 x^{2} / 8 d x=\int_{0}^{2} 3 x^{3} / 8 d x=\left.\frac{3}{8} \cdot \frac{x^{4}}{4}\right|_{0} ^{2}=1.5
$$ | 1.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,016 |
Example 11. Find the mathematical expectation of the random variable $X$, if the distribution function of this variable is known
$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x^{2} & \text { if } 0 < x \leq 1\end{cases}
$$ | Solution. First, let's find the probability density function of this variable. Using formula (2.3.5), we get
$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 2 x & \text { if } 0 < x \leq 1 \\ 0 & \text { if } x > 1\end{cases}
$$
Therefore,
$$
M(X)=\int_{0}^{1} x \cdot 2 x d x=\int_{0}^{1} 2 x^{2} d x=2 \int_{0}^{... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,017 |
Example 13. Find the mathematical expectation of a random variable $X$, the distribution function of which is given by
$$
F(x)= \begin{cases}0 & \text { if } x \leq -a \\ \frac{(a+x)^{2}}{2 a^{2}} & \text { if } -a < x \leq a \\ 1 & \text { if } x > a\end{cases}
$$ | Solution. First, let's find the probability density function of this random variable. Since \( p(x) = F'(x) \), we have:
\[
p(x)= \begin{cases}0 & \text{if } x \leq -a \\ \frac{1}{a}\left(1+\frac{x}{a}\right) & \text{if } -a < x \leq 0 \\ \frac{1}{a}\left(1-\frac{x}{a}\right) & \text{if } 0 < x \leq a \\ 0 & \text{if ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,018 |
Example 4. The distribution law of a discrete random variable $X$ is given by the table
| $X$ | -2 | -1 | 0 | 1 | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.2 | 0.4 | 0.2 | 0.1 |
Calculate the variance of the random variable $X$ using formula (2.5.4) and formula (2.5.10). | Solution. First, we find the mathematical expectation of the random variable $X$:
$$
M(X)=-2 \cdot 0.1-1 \cdot 0.2+0 \cdot 0.4+1 \cdot 0.2+2 \cdot 0.1=0
$$
Next, we determine the distribution law of the random variable $(X-M(X))^{2}$:
| $(X-M(X))^{2}$ | $(-2-0)^{2}$ | $(-1-0)^{2}$ | $(0-0)^{2}$ | $(1-0)^{2}$ | $(2-0... | 1.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,021 |
Example 5. A symmetric coin is tossed 4 times. The random variable $X$ - "the number of heads appearing in these tosses". Find the numerical characteristics of the random variable $X: M(X), D(X), \sigma(X)$. | Solution. The discrete random variable $X$ can take five values: $x_{1}=0, x_{2}=1, x_{3}=2, x_{4}=3, x_{5}=4$.
The distribution law of the random variable $X$ can be given by the table
| $X$ | 0 | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | $1 / 16$ | $4 / 16$ | $6 / 16$ | $4 / 16$ | $1... | M(X)=2,D(X)=1,\sigma(X)=1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,022 |
Example 6. Find the variance of the discrete random variable $X$ representing the number of points that fall when tossing a die. | Solution. First, let's write down the distribution law of this random variable in the form of a table
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |
Let's find the mathema... | \frac{35}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,023 |
Example 7. Given all possible values of a discrete random variable $X: x_{1}=1, x_{2}=2, x_{3}=3$, and it is known that $M(X)=2.3$, $M\left(X^{2}\right)=5.9$. Find the distribution law of the random variable $X$. | Solution. Let's write down the distribution laws of the discrete random variables $X$ and $X^{2}$:
| $X$ | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: |
| $P$ | $p_{1}$ | $p_{2}$ | $p_{3}$ |
| $X^{2}$ | 1 | 4 | 9 |
| :---: | :---: | :---: | :---: |
| $P$ | $p_{1}$ | $p_{2}$ | $p_{3}$ |,
where $p_{1}, p_{2}, p_{3}$ ... | \begin{pmatrix}X&1&2&3\\\hlineP&0.2&0.3&0.5\\\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,024 |
Example 8. A discrete random variable $X$ can take only two values $x_{1}$ and $x_{2}$, with $x_{1}<x_{2}$. The probability $p_{1}=0.5$, the expected value $M(X)=3.5$, and the variance $D(X)=0.25$ are known. Find the distribution law of the discrete random variable $X$. | Solution. Since $p_{1}+p_{2}=1$ (see formula (2.1.2)) and $p_{1}=0.5$, then $p_{2}=0.5 ; M(X)=0.5 x_{1}+0.5 x_{2}=3.5$, from which $x_{1}+x_{2}=7$. By formula (2.5.12) we find
$D(X)=\left(x_{1}-3.5\right)^{2} p_{1}+\left(x_{2}-3.5\right)^{2} p_{2}=\left(x_{1}-3.5\right)^{2} \cdot 0.5+\left(x_{2}-3.5\right)^{2} \cdot 0... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,025 | |
Example 10. Find the numerical characteristics $M(X), D(\dot{X}), \sigma(X)$ of a continuous random variable $X$ given by the probability density function
$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 2 x & \text { if } 0 < x \leq 1 \\ 0 & \text { if } x > 1\end{cases}
$$ | Solution. Using formula (2.4.7), we find the expected value:
$$
M(X)=\int_{0}^{1} 2 x \cdot x d x=\int_{0}^{1} 2 x^{2} d x=\left.\frac{2}{3} \cdot x^{3}\right|_{0} ^{1}=\frac{2}{3}
$$
Using formulas (2.5.13) and (2.5.16) respectively, we obtain
$$
\begin{gathered}
D(X)=\int_{0}^{1}\left(x-\frac{2}{3}\right)^{2} \cdo... | M(X)=\frac{2}{3},D(X)=\frac{1}{18},\sigma(X)\approx0.24 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,026 |
Example 11. The random variable $X$ is given by the distribution function
$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x^{3} & \text { if } 0 < x \leq 1 \\ 1 & \text { if } x > 1\end{cases}
$$
Find the numerical characteristics of the random variable $X: M(X), D(X), \sigma(X)$ | Solution. First, we find the probability density function $p(x)$ using formula (2.3.5). Since $p(x)=F^{\prime}(x)$, we have
$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 3 x^{2} & \text { if } 0 < x \leq 1\end{cases}
$$
Using formula (2.4.7), we calculate the expected value:
$$
M(X)=\int_{0}^{1} x \cdot 3 x^{2}... | M(X)=\frac{3}{4},D(X)=\frac{3}{80},\sigma(X)\approx0.19 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,027 |
Example 1. Prove that the initial moment of zero order is equal to one, and the initial moment of the first order of a random variable $X$ is equal to its mathematical expectation. | Solution. Formulas (2.6.2), (2.6.4), (2.6.6) for $k=0$ take the following forms respectively:
$$
\begin{aligned}
& v_{0}=\sum_{i=1}^{n} x_{i}^{0} p_{i}=\sum_{i=1}^{n} p_{i}=1 \\
& v_{0}=\sum_{i=1}^{\infty} x_{i}^{0} p_{i}=\sum_{i=1}^{\infty} p_{i}=1 \\
& v_{0}=\int_{-\infty}^{+\infty} x^{0} p(x) d x=\int_{-\infty}^{+\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,028 |
Example 2. Prove that the central moment of zero order is equal to one; the central moment of the first order is equal to zero; the central moment of the second order of a random variable $X$ is equal to the variance of this variable. | Solution. When $k=0$, formulas (2.6.9), (2.6.10), (2.6.11) take the following forms respectively:
$$
\mu_{0}=\sum_{i=1}^{n}\left(x_{1}-a\right)^{0} p_{i}=\sum_{i=1}^{n} p_{i}=1
$$
$$
\begin{gathered}
\mu_{0}=\sum_{i=1}^{\infty}\left(x_{i}-a\right)^{0} p_{i}=\sum_{i=1}^{\infty} p_{i}=1 \\
\mu_{0}=\int_{-\infty}^{+\inf... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,029 |
Example 3. Express the central moments of the second, third, and fourth orders through the initial moments. | Solution. Taking into account the properties of the expectation, the definitions of moments, and assuming \( M(X) = a \), we obtain
\[
\begin{aligned}
& \mu_{2}=M(X-a)^{2}=M\left(X^{2}-2 a X+a^{2}\right)=M\left(X^{2}\right)-2 a M(X)+M\left(a^{2}\right)= \\
& =M\left(X^{2}\right)-2 a^{2}+a^{2}=M\left(X^{2}\right)-a^{2}... | \begin{aligned}&\mu_{2}=v_{2}-v_{1}^{2},\\&\mu_{3}=v_{3}-3v_{1}v_{2}+2v_{1}^{3},\\&\mu_{4}=v_{4}-4v_{1}v_{3}+6v_{1}^{2}v_{2}- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,030 |
Example 4. A discrete random variable $X$ is given by the distribution law:
| $X$ | 1 | 2 |
| :---: | :---: | :---: |
| $P$ | 0.4 | 0.6 |
Find the first, second, and third central moments of the random variable $X$. | Solution. According to the definition (see formula 2.6.I), we first find the initial moments using formula (2.6.2):
$$
\begin{gathered}
v_{1}=M(X)=1 \cdot 0.4 + 2 \cdot 0.6 = 1.6, \quad(a=1.6) \\
v_{2}=M\left(X^{2}\right)=1^{2} \cdot 0.4 + 2^{2} \cdot 0.6 = 2.8 \\
v_{3}=M\left(X^{3}\right)=1^{3} \cdot 0.4 + 2^{3} \cdo... | \mu_{1}=0,\mu_{2}=0.24,\mu_{3}=-0.48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,031 |
Example 5. Find the first, second, and third order moments of the random variable $X$ with the probability density function
$$
p(x)=\left\{\begin{array}{lr}
0 & \text { if } x \leq 0 \\
e^{-x} & \text { if } x>0
\end{array}\right.
$$ | Solution. Let's find the initial moments according to formula (2.6.6):
$$
\begin{aligned}
& v_{1}=M(X)=\int_{-\infty}^{+\infty} x p(x) d x=\int_{0}^{+\infty} x e^{-x} d x=\int_{0}^{+\infty} x d\left(-e^{-x}\right)= \\
& =-\left.x e^{-x}\right|_{0} ^{+\infty}+\int_{0}^{+\infty} e^{-x} d x=0-\left.e^{-x}\right|_{0} ^{+\... | v_{1}=1,v_{2}=2,v_{3}=6;\mu_{1}=0,\mu_{2}=1,\mu_{3}=2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,032 |
Example 6. Find the initial moments of the random variable $X$ with the probability density function
$$
p(x)= \begin{cases}0 & \text { if } x \leq 1 \\ \frac{5}{x^{6}} & \text { if } x>1\end{cases}
$$ | Solution. According to formula (2.6.6), we obtain
$$
v_{k}=M\left(X^{k}\right)=\int_{-\infty}^{+\infty} x^{k} \frac{5}{x^{6}} d x=5 \int_{1}^{+\infty} x^{k-6} d x=\left.5 \frac{x^{k-5}}{k-5}\right|_{1} ^{+\infty}
$$
The improper integral converges for $k<5$. In this case,
$$
v_{k}=M\left(X^{k}\right)=\left.5 \frac{x... | v_{k}=\frac{5}{5-k} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,033 |
Example 7. A random variable $X$ is given by the distribution function
$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x^{2} & \text { if } 0 < x \leq 1 \\ 1 & \text { if } x > 1\end{cases}
$$
Find the initial and central moments of the first three orders of the random variable $X$. | Solution. First, we find the probability density function of the given random variable $X$. Since $p(x) = F'(x)$, the probability density function of this random variable is defined by the function
$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 2 x & \text { if } 0 < x \leq 1 \\ 0 & \text { if } x > 1\end{cases}
$... | \mu_1=0,\mu_2=\frac{1}{18},\mu_3=-\frac{1}{135} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,034 |
Example 1. A discrete random variable $X$ is given by the distribution law
| $X$ | 1 | 2 | 3 | 4 | 5 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.2 | 0.4 | 0.2 | 0.1 |
Write the distribution law of the random variable $Y=X^{3}$. | Solution. The random variable $Y$ takes values equal to the cubes of the values of the variable $X$, with the same probabilities as the random variable $X$. The cubes of the values of the random variable $X$ are:
$$
\begin{gathered}
y_{1}=x_{1}^{3}=1, y_{2}=x_{2}^{3}=8, y_{3}=x_{3}^{3}=3^{3}=27 \\
y_{4}=x_{4}^{3}=4^{3... | \begin{pmatrix}\hlineY=X^{3}&1&8&27&64&125\\\hlineP&0.1&0.2&0.4&0.2&0.1\\\hline\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,035 |
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