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Example 2. A discrete random variable $X$ is given by the distribution law
| $X$ | -2 | -1 | 0 | 1 | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.2 | 0.15 | 0.25 | 0.3 |
Find the distribution law and the mathematical expectation of the random variable $Y=X^{2}$. | Solution. The square of a random variable $X$, i.e., $Y=X^{2}$, is a new random variable that takes values equal to the squares of its values with the same probabilities as the random variable $X$.
The squares of the random variable $X$ are: $(-2)^{2}=4,(-1)^{2}=1$, $0^{2}=0,1^{2}=1,2^{2}=4$, i.e., the random variable... | 2.05 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,036 |
Example 3. A discrete random variable $X$ has the distribution law
| $X$ | 0 | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.3 | 0.4 | 0.2 |
Find the distribution law of the random variable $Y=\sin \frac{\pi}{2} X+1$. | Solution. First, let's find the values of the function $Y=f(X)=\sin \frac{\pi}{2} X+1$.
For $x=0,1,2,3$, we get the numbers $1,2,1,0$ respectively. Therefore, the possible values of the random variable $Y$ are $y_{1}=0, y_{2}=1, y_{3}=2$. Let's calculate the probabilities of these values:
$$
\begin{aligned}
& P(Y=0)=... | \begin{pmatrix}Y=\sin\frac{\pi}{2}X+1&0&1&2\\\hlineP&0.2&0.5&0.3\\\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,037 |
Example 4. The random variable $X$ is given by the probability density function
$$
p(x)= \begin{cases}1 & \text { if } 1 \leq x \leq 2 \\ 0 & \text { if } x < 1 \text{ or } x > 2\end{cases}
$$
Find the probability density function of the function $Y=X^{2}$. | Solution. On the interval of possible values of the random variable $X$, the function $y=x^{2}$ is monotonically increasing. The inverse function $x=\sqrt{y}$ is also monotonically increasing on the interval $[1 ; 4]$ - the range of possible values of the random variable $Y$. We find the derivative of the inverse funct... | (y)=\begin{cases}1/2\sqrt{y}&\text{for}1\leqy\leq4\\0&\text{for}y>4\end{cases} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,038 |
Example 5. The random variable $X$ is given by the probability density function
$$
p(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-x^{2} / 2 \sigma^{2}}
$$
Find the probability density function of $Y=X^{3}$. | Solution. Since the function $y=x^{3}$ is differentiable and strictly monotonic, we can apply formula (2.7.4).
We find the function $x=\varphi(y)$, which is the inverse of the function $y=x^{3}$: $\varphi(y)=x=y^{1 / 3}$ and $p[\varphi(y)]$. According to the condition, $p(x)$ is defined by formula (I), so
$$
p[\varph... | (y)=\frac{1}{3\sigmay^{2/3}\sqrt{2\pi}}e^{-y^{2/3}/2\sigma^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,039 |
Example 6. A continuous random variable $X$ is given by the probability density function $p(x)=\cos x$ in the interval $(0 ; \pi / 2)$; outside this interval, $p(x)=0$. Find the expected value of the function $Y=X^{2}$. | Solution. We will use formula (2.7.7). In this case, $p(x)=\cos x, f(x)=x^{2}, a=0, b=\pi / 2$. Therefore,
$$
\begin{gathered}
M\left(X^{2}\right)=\int_{0}^{\pi / 2} x^{2} \cos x d x=\int_{0}^{\pi / 2} x^{2} d(\sin x)=\left.x^{2} \sin x\right|_{0} ^{\pi / 2}-2 \int_{0}^{\pi / 2} x \sin x d x=\frac{\pi^{2}}{4}+ \\
+2 \... | \frac{\pi^{2}}{4}-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,040 |
Example 7. Discrete independent random variables are given by the distribution laws:
| $X$ | 5 | 6 |
| :---: | :---: | :---: |
| $P$ | 0.4 | 0.6 |
| $Y$ | 7 | 8 |
| :---: | :---: | :---: |
| $P$ | 0.8 | 0.2 |
Formulate the distribution law of the random variable $Z=X+Y$. | Solution. The possible values of the quantity $Z$ are the sums of each possible value of $X$ with all possible values of $Y$:
$$
z_{1}=5+7=12, z_{2}=5+8=13, z_{3}=6+7=13, z_{4}=6+8=14
$$
Let's find the probabilities of these possible values. For $Z=12$, it is sufficient for the quantity $X$ to take the value $x_{1}=5... | \begin{pmatrix}Z&12&13&14\\\hlineP&0.32&0.56&0.12\\\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,041 |
Example 8. Independent random variables $X$ and $Y$ are given by the probability density functions
\[
\begin{aligned}
& p(x)=\frac{1}{5} e^{-x / 5} \quad(0 \leq x<+\infty) \\
& p(y)=\frac{1}{6} e^{-y / 6} \quad(0 \leq y<+\infty)
\end{aligned}
\]
Find the composition of these laws, i.e., the probability density functi... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,042 | ||
Example 2. The distribution law of a discrete two-dimensional random variable is given by the table
Find the conditional distribution law of the variable $Y$ given $X=1$. Are the variables $X$ and $Y$ independent? | Solution. The probabilities of the values of the quantity $Y$ when $X=1$ will be found using formula (2.8.5).
Since
$$
P(X=1)=0.15+0.20+0.12+0.03=0.5
$$
then by formula (2.8.5) we get
$$
\begin{gathered}
P(Y=1 / X=1)=\frac{0.15}{0.5}=0.3, P(Y=2 / X=1)=\frac{0.2}{0.5}=0.4 \\
P(Y=3 / X=1)=\frac{0.12}{0.5}=0.24, P(Y=4... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,043 |
Example 4. Find the probability density $p(x, y)$ of a two-dimensional random variable $(X, Y)$ given the known distribution function
$$
F(x, y)=\left(\frac{1}{\pi} \operatorname{arctg} x+\frac{1}{2}\right) \cdot\left(\frac{1}{\pi} \operatorname{arctg} y+\frac{1}{2}\right)
$$ | ## Solution. Since
$$
\begin{gathered}
F_{x}^{\prime}(x, y)=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}} \cdot\left(\frac{1}{\pi} \operatorname{arctg} y+\frac{1}{2}\right) \\
F_{x y}^{\prime \prime}(x, y)=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}} \cdot \frac{1}{\pi} \cdot \frac{1}{1+y^{2}}
\end{gathered}
$$
then by formula (2.8... | p(x,y)=\frac{1}{\pi^{2}(1+x^{2})(1+y^{2})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,044 |
Example 5. Given the probability density function of a two-dimensional random variable $(X, Y)$
$$
\begin{aligned}
& p(x, y)=0.5 \sin (x+y) \quad\left(0 \leq x \leq \frac{\pi}{2}, 0 \leq y \leq \frac{\pi}{2}\right) . \\
& p(x, y)=0 \quad \text { outside the square } \quad S=\left[0, \frac{\pi}{2} ; 0, \frac{\pi}{2}\ri... | Solution. By formula (2.8.12) we get
$$
\begin{aligned}
& F(x, y)=\int_{0}^{x} \int_{0}^{y} 0.5 \sin (u+v) d u d v=\int_{0}^{x}\left[\int_{0}^{y} 0.5 \sin (u+v) d v\right] d u= \\
& \int_{0}^{x}\left(-0.5 \cos (u+v)\right|_{v=0} ^{v=y} d u=-0.5 \int_{0}^{x}[\cos (u+y)-\cos u] d u= \\
& =-0.5 \int_{0}^{x} \cos (u+y) d ... | 0.5[\sin\frac{\pi}{6}+\sin\frac{\pi}{6}-\sin\frac{\pi}{3}]\approx0.67 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,045 |
Example 6. Independent random variables $X$ and $Y$ have the following densities:
$$
\begin{aligned}
& p_{1}(x)= \begin{cases}0 & \text { if } x1, \\
0.5 & \text { if }-12 \\
0.5 & \text { if } 0<y \leq 2 .\end{cases}
\end{aligned}
$$
Find: 1) the distribution functions $F_{1}(x)$ and $F_{2}(y)$;
2) the density of t... | Solution. The distribution functions $F_{1}(x)$ and $F_{2}(y)$ will be found using formula (2.3.2)
$$
\text { If } x \leq-1, \text { then } \int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{x} 0 d t=0 ; F_{1}(x)=0
$$
If $-1 < x \leq 1$
$$
F_{1}(x)=\int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{-1} 0 d t+\int_{-1}^{x} 0.5 d t... | \begin{aligned}&F(x,y)=\begin{cases}0&\text{if}x\leq-1\text{}y\leq0\\0.25(x+1)y&\text{if}-1<x\leq1,0<y\leq2\\0.5y&\text{if}x>1,0<y | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,046 |
Example 1. The probability of hitting the target with one shot for a given shooter is 0.7 and does not depend on the shot number. Find the probability that exactly 2 hits will occur out of 5 shots. | Solution. Since $p=0.7$, then $q=1-p=1-0.7=0.3$. According to the condition $n=5, k=2$, using formula (3.1.1) we find
$$
P_{5}(2)=C_{5}^{2} p^{2} q^{5-2}=10 \cdot(0.7)^{2} \cdot(0.3)^{3}=10 \cdot 0.49 \cdot 0.027=0.1323
$$ | 0.1323 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,047 |
Example 2. Five fair coins are tossed. Find the probability that: exactly 2 heads appeared; more than one head appeared. | Solution. Let $X$ be the number of heads that appear in these tosses. In this case, $p=1 / 2$ and $q=1 / 2$.
$$
P(X=2)=P_{5}(2)=C_{5}^{2} p^{2} q^{s-2}=10 \cdot\left(\frac{1}{2}\right)^{2} \cdot\left(\frac{1}{2}\right)^{3}=10 \cdot \frac{1}{4} \cdot \frac{1}{8}=\frac{5}{16}=0.3125
$$
The random variable $X$ can take ... | 0.8125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,048 |
Example 3. The germination rate of seeds of this plant is $90 \%$. Find the probability that out of four sown seeds, the following will germinate: a). three; b) no less than three. | Solution. The required probabilities are found using the Bernoulli formula.
In the first case $n=4, k=3, p=0.9, q=1-p=0.1$, therefore
$$
P_{4}(3)=C_{4}^{3} p^{3} q=4 \cdot 0.729 \cdot 0.1=0.2916
$$
In the second case, event $A$ consists of either three or four seeds sprouting out of four. By the addition theorem of ... | 0.9477 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,049 |
Pr i mer 7. For the normal operation of the car base, there should be no less than eight cars on the line, and there are ten of them. The probability of each car not coming out on the line is 0.1. Find the probability of the normal operation of the car base for the next day. | Solution. The car base will work normally (event $D$) if eight (event $A$), nine (event $B$), or all ten (event $C$) cars go out on the line. According to the theorem of the addition of probabilities,
$$
P(D)=P(A)+P(B)+P(C)=P_{10}(8)+P_{10}(9)+P_{10}(10)
$$
Each term will be found using the Bernoulli formula.
Since ... | 0.9298 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,050 |
Example 8. The average germination rate of seeds is $80 \%$. Find the most probable number of germinated seeds among nine seeds. | Solution. The number $k_{0}$ will be determined using inequalities (3.1.5). Since $n=9, p=0.8$, and $q=0.2$, then $9 \cdot 0.8 - 0.2 \leq k_{0} \leq 9 \cdot 0.8 + 0.8 = 8$. An integer is obtained; hence, there are two most probable numbers of germinated seeds: 8 and 7. Their probabilities are the highest and equal to e... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,051 |
Example 9. A coin is tossed 10 times. Find the probability that heads will appear: a) from 4 to 6 times; b) at least once. | Solution. According to formula (3.1.6) for $n=10, k_{1}=4, k_{2}=6, p=q=0.5$ we find
$$
P_{10}(4 \leq k \leq 6)=P_{10}(4)+P_{10}(5)+P_{10}(6)=\frac{210}{1024}+\frac{252}{1024}+\frac{210}{1024}=\frac{21}{32}
$$
According to formula (3.1.7) we obtain
$$
P_{10}(1 \leq k \leq 10)=1-\left(\frac{1}{2}\right)^{10}=\frac{10... | \frac{21}{32} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,052 |
Example 10. How many times should a die be thrown so that the most probable number of times a two appears is 32? | Solution. In this case, $k_{0}=32, p=\frac{1}{6}, q=\frac{5}{6}$. The inequalities $n p-q \leq k_{0} \leq n p+p$ become
$$
n \cdot \frac{1}{6}-\frac{5}{6} \leq 32 \leq n \cdot \frac{1}{6}+\frac{1}{6}
$$
or
$$
n \cdot \frac{1}{6}-\frac{5}{6} \leq 32 ; \quad n \cdot \frac{1}{6}+\frac{1}{6} \geq 32
$$
From the first i... | 191\leqn\leq197 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,053 |
Example 11. What is the probability of event $A$ occurring in each trial, if the most probable number of occurrences of event $A$ in 120 trials is 32? | Solution. Given $n=120, k_{0}=32$. Inequalities (3.1.5) take the form
$$
120 p-(1-p) \leq 32 \leq 120 p+p
$$
or
$$
120 p-(1-p) \leq 32, \quad 120 p+p \geq 32
$$
Solving this system of inequalities, we find that
$$
\frac{32}{121} \leq p \leq \frac{33}{121}
$$ | \frac{32}{121}\leqp\leq\frac{33}{121} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,054 |
Example 12. What is the minimum number of trials that need to be conducted so that with a probability of at least $\alpha(0<\alpha<1)$, one can expect the event $A$ to occur at least once, if the probability of event $A$ in one trial is $p$? | Solution. We require that the probability of event $A$ occurring at least once in $n$ trials (see formula (3.1.7)) is not less than $\alpha$:
$$
1-q^{n} \geq \alpha, \text { or }\left(1-(1-p)^{n}\right) \geq \alpha \text {. }
$$
Solving this inequality for $n$, we obtain the inequality
$$
n \geq \frac{\lg (1-\alpha)... | n_{0}=[\frac{\lg(1-\alpha)}{\lg(1-p)}]+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,055 |
Example 13. The target consists of 3 pairwise non-intersecting zones. The probability of hitting the first zone with one shot for a given shooter is 0.5. For the second and third zones, this probability is 0.3 and 0.2, respectively. The shooter fires 6 shots at the target. Find the probability that there will be 3 hits... | Solution. To find the required probability, we will use formula (3.1.12).
Since in this case $n=6, k_{1}=3, k_{2}=2, k_{3}=1, p_{1}=0.5$, $p_{2}=0.3, p_{3}=0.2$, then
$$
P_{6}(3,2,1)=\frac{6!}{3!2!1!}(0.5)^{3} \cdot(0.3)^{2} \cdot 0.2=60 \cdot 0.125 \cdot 0.09 \cdot 0.2=0.135
$$
Example 14. A treatment method leads ... | 0.135 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,056 |
Example 15. Two equally strong chess players are playing chess. What is more likely: a) to win one game out of two or two games out of four? b) to win at least two games out of four or at least three games out of five? Draws are not considered. | Solution. Since the chess players are of equal strength, the probability of winning $p=1 / 2$; therefore, the probability of losing $q$ is also
equal to $1 / 2$. In all games, the probability of winning is constant and it does not matter in what sequence the games will be won, so the Bernoulli formula can be applied.
... | \frac{11}{16}>\frac{8}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,057 |
Example 16. The quality check of the produced parts showed that on average, the defect rate is $7.5 \%$. Find the most probable number of standard parts in a batch of 39 pieces, randomly selected.
翻译结果如上，保留了原文的换行和格式。 | Solution. Extracting a standard part (event $A$) and extracting a non-standard part (event $\bar{A}$) are opposite events. From the condition, it follows that $q=0.075$, so $p=1-0.075=0.925$. Since in this case $n=39$, the formula (3.1.5) for determining the most probable number of occurrences of event $A$ takes the fo... | 36or37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,058 |
Example 17. When shooting at a target, the probability of hitting the target with one shot is 0.7. For what number of shots is the most probable number of hits equal to 16? | Solution. We refer to formula (3.1.5). In this case, $p=0.7$, $q=0.3$, $k_{0}=16$; the number $n$ is unknown and needs to be found. Formula $(3.1 .5)$ takes the form
$$
0.7 n-0.3 \leq 16 \leq 0.7 n+0.7
$$
From these inequalities, it follows that
$$
\begin{aligned}
& 0.7 n \leq 16.3, \quad 7 n \leq 163, \quad n \leq ... | 22or23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,059 |
Example 19. A coin is tossed 5 times. The random variable $X$ is the number of times the head appears. Possible values of the variable $X: \cdot x_{0}=0$, $x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=5$. Write the distribution law of the random variable $X$. | Solution. The probabilities of the specified values $P\left(x=x_{k}\right)$ will be found using the Bernoulli formula, taking into account that in this case $p=1 / 2, q=1 / 2$.
$$
\begin{gathered}
P_{5}(0)=C_{n}^{0} p^{0} q^{n-0}=C_{5}^{0} \cdot\left(\frac{1}{2}\right)^{0} \cdot\left(\frac{1}{2}\right)^{5}=\frac{1}{2^... | \begin{pmatrix}X&0&1&2&3&4&5\\\hlineP&\frac{1}{32}&\frac{5}{32}&\frac{10}{32}&\frac{10}{32}&\frac{5}{32}&\frac{1}{32}\end{pmatrix} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,060 |
Example 20. Will the probability $P_{n}(k)$, determined by the Bernoulli formula, increase or decrease if two more trials are added to the total number of trials, while the probability $p$ of event $A$ occurring in a single trial remains unchanged? | Solution. Since
$$
P_{n}(k)=C_{n}^{k} p^{k} q^{n-k}=\frac{n!}{k!(n-k)!} p^{k} q^{n-k}
$$
then
$$
P_{n+2}(k)=C_{n+2}^{k} p^{k} q^{n+2-k}=\frac{(n+2)!}{k!(n+2-k)!} p^{k} q^{n+2-k}
$$
Let's find the ratio of $P_{n+2}(k)$ to $P_{n}(k)$:
$$
\begin{aligned}
& \frac{P_{n+2}(k)}{P_{n}(k)}=\frac{(n+2)!}{k!(n+2-k)!} p^{k} q... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,061 |
Example 2. Find the variance of a discrete random variable distributed according to the binomial law. | Solution. The random variable $X$ - "the number of occurrences of event $A$ in $n$ trials" can be represented as the sum $X=X_{1}+X_{2}+\ldots+X_{n}$, where $X_{k}$ is "the number of occurrences of event $A$ in the $k$-th trial $(k=1,2, \ldots, n)$". According to formula (2.5.9), we have
$$
D(X)=D\left(X_{1}\right)+D\... | D(X)=npq | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,062 |
Example 3. Find the root mean square deviation of a random variable $X$, distributed according to the binomial law. | Solution. Taking into account the definition of the standard deviation (see formula (2.5.16)) and formula (3.2.6) (see example 2), we obtain
$$
G(X)=\sqrt{n p q}
$$ | G(X)=\sqrt{npq} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,063 |
Example 4. The probability of hitting the target with a single shot is $p=0.8$. Find the probability of five hits in six shots. | Solution. We apply formula (3.2.1). Since by condition $n=6$, $k=5$, $p=0.8$, therefore, $q=1-p=0.2$, then
$$
\begin{gathered}
P_{6}(5)=C_{6}^{5} p^{5} q^{1}=\frac{6!}{5!} \cdot(0.8)^{5} \cdot(0.2)^{1}=6 \cdot(0.8)^{5} \cdot 0.2= \\
=6 \cdot 0.32768 \cdot 0.2=0.3934 .
\end{gathered}
$$ | 0.3934 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,064 |
Example 5. The germination rate of seeds of this variety of plants is $80 \%$. What is the probability that out of 5 sown seeds, at least 4 will germinate. | Solution. According to the condition $p=0.8$, therefore $q=0.2$. Further, $n=5, k \geq 4$, i.e., $k$ takes the values 4 or 5. Let the event whose probability we need to determine be denoted by $A$, then $P(A)=P_{5}(4)+P_{5}(5)$. According to the Bernoulli formula, we find the probabilities $P_{5}(4), P_{5}(5)$ and thei... | 0.73728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,065 |
Example 7. Quality control has established that, on average, 75 out of every 100 parts are not defective. Construct the binomial probability distribution for the number of suitable parts out of 6 parts taken at random. | Solution. From the condition of the problem, it follows that $p=0.75, q=0.25, n=6$. According to the Bernoulli formula, we find:
$$
\begin{aligned}
& P_{6}(0)=1 \cdot(0.25)^{6} \approx 0.0002 ; \quad P_{6}(1)=6 \cdot(0.75) \cdot(0.25)^{5} \approx 0.004 ; \\
& P_{6}(2)=15 \cdot(0.75)^{2} \cdot(0.25)^{4} \approx 0.033 ;... | 0.004+0.033+0.132+0.297+0.356+0.178=1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,066 |
Example 8. 9 independent trials are conducted. In each trial, event $A$ occurs with the same probability $p=2 / 3$. Write the distribution law of the random variable $X$ "the number of occurrences of event $A$ in these trials" in the form of a table. | Solution. Using the Bernoulli formula, we calculate the probabilities $P_{9}(k)$, where $k=0,1,2,3,4,5,6,7,8,9$:
$$
\begin{aligned}
& P_{9}(0)=C_{9}^{0} p^{0} q^{9}=\left(\frac{1}{3}\right)^{9}=0.0000 \\
& P_{9}(1)=C_{9}^{1} p^{1} q^{8}=9 \cdot \frac{2}{3} \cdot\left(\frac{1}{3}\right)^{8}=0.0009 \\
& P_{9}(2)=C_{9}^{... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,067 |
Example 9. Prove the recurrence formula for binomial probabilities:
$$
P_{n}(k+1)=\frac{p}{q} \frac{n-k}{k+1} P_{n}(k)
$$ | Solution. Since
$$
P_{n}(k)=C_{n}^{k} p^{k} q^{n-k}=\frac{n!}{k!(n-k)!} p^{k} q^{n-k}
$$
then
$$
P_{n}(k+1)=C_{n}^{k+1} p^{k+1} q^{n-(k+1)}=\frac{n!}{(k+1)!(n-(k+1))!} p^{k+1} q^{n-k-1}
$$
Transform the last expression:
$$
\begin{gathered}
P_{n}(k+1)=\frac{1}{k+1} \cdot \frac{n!}{k!(n-k-1)!} \cdot p \cdot p^{k} \c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,068 |
Example 1. Prove that the Poisson distribution is a limiting case of the binomial distribution. | Solution. From equality (3.3.2), we determine $p$ and $q$:
$$
p=\frac{a}{n}, \quad q=1-\frac{a}{n}
$$
and substitute into the Bernoulli formula:
$$
\begin{aligned}
P_{n}(k) & =C_{n}^{k} p^{k} q^{n-k}=\frac{n(n-1) \ldots(n-(k-1))}{k!} \cdot\left(\frac{a}{n}\right)^{k}\left(1-\frac{a}{n}\right)^{n-k}= \\
& =\frac{a^{k... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,069 |
Example 2. Prove that
$$
\sum_{k=0}^{\infty} p_{k}=\sum_{k=0}^{\infty} \frac{a^{k} e^{-a}}{k!}=1
$$ | Solution. Taking into account the power series expansion of the function $f(x)=e^{x}$
$$
e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots+\frac{x^{k}}{k!}+\ldots
$$
and the resulting equality
$$
e^{a}=1+\frac{a}{1!}+\frac{a^{2}}{2!}+\frac{a^{3}}{3!}+\ldots+\frac{a^{k}}{k!}+\ldots
$$
we obtain
$$
\sum_... | 1 | Algebra | proof | Yes | Yes | olympiads | false | 33,070 |
Example 3. Find the mathematical expectation of a discrete random variable distributed according to the Poisson law. | Solution. Based on the table defining the distribution law and formula (2.4.6), we find
$$
\begin{gathered}
M(X)=0 \cdot e^{-a}+1 \cdot \frac{a}{1!} e^{-a}+2 \cdot \frac{a^{2}}{2!} e^{-a}+\ldots+k \frac{a^{k}}{k!} e^{-a}+\ldots= \\
=\sum_{k=1}^{\infty} \frac{a^{k}}{(k-1)!} e^{-a}=a e^{-a} \sum_{k=1}^{\infty} \frac{a^{... | D(X)= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,071 |
Example 5. Prove that the sum of two independent random variables, distributed according to the Poisson law with parameters $a$ and $b$, is also distributed according to the Poisson law with parameter $a+b$. | Solution. By the condition for random variables $X$ and $Y$
$$
P(X=k)=\frac{a^{k}}{k!} e^{-a}, \quad P(Y=k)=\frac{b^{k}}{k!} e^{-b}
$$
Since random variables $X$ and $Y$ are independent, the random events $(X=m)$ and $(Y=k-m)$ are independent for any non-negative integers $k$ and $m, 0 \leq m \leq k$, therefore
$$
\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,072 |
Example 6. A random variable $X$ is distributed according to the Poisson law. Find $P(X=3)$, if $a=4$, as well as the mathematical expectation and variance of the variable $X$. | Solution. Using formula (3.3.1), we find
$$
P(X=3)=\frac{4^{3}}{3!} e^{-4}=\frac{32}{3} \cdot \frac{1}{e^{4}}=10.6667 \cdot 0.0183=0.1952
$$
According to formulas (3.3.5) and (3.3.6), we obtain
$$
M(X)=4, \quad D(X)=4
$$ | P(X=3)=0.1952,\quadM(X)=4,\quadD(X)=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,073 |
Example 9. The probability of manufacturing a non-standard part is $p=0.004$. Find the probability that among 1000 parts there will be 5 non-standard ones. | Solution. Here $n=1000, p=0.004, a=n p=1000 \cdot 0.004=4$. Using formula (3.3.1), we find
$$
P_{1000}(5)=\frac{4^{5}}{5!} e^{-4}=\frac{128}{15} \cdot \frac{1}{e^{4}}=8.5333 \cdot 0.0183=0.1562
$$
Example 10. A spinner operates 1000 spindles. The probability of a thread break on one spindle within 1 minute is 0.002. ... | 0.1428 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,074 |
Example 11. Independent trials are conducted, in each of which event $A$ can occur with a probability of 0.001. What is the probability that in 2000 trials, event $A$ will occur no less than two and no more than four times. | Solution. From the condition of the problem, it follows that $n=2000, p=0.001$, $a=n p=2000 \cdot 0.001=2, 2 \leq k \leq 4$. Therefore,
$$
P_{n}(2 \leq k \leq 4)=P_{n}(2)+P_{n}(3)+P_{n}(4)=\frac{2^{2}}{2!} e^{-2}+\frac{2^{3}}{3!} e^{-2}+\frac{2^{4}}{4!} e^{-2} \approx 0.541
$$ | 0.541 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,075 |
Example 12. A factory sent 5000 good-quality items to a base. The probability that an item will be damaged during transportation is 0.0002. What is the probability that 3 defective items will arrive at the base?
$\mathrm{Pe} \mathrm{sh} \mathbf{e n i e . ~ F r o m ~ t h e ~ c o n d i t i o n ~ i t ~ f o l l o w s ~ t ... | Solution. Here it is required to find the probabilities: 1) $P_{1000}(2)$;
2) $P_{1000}(k \geq 2)$. According to the condition, $n=1000, p=0.001, a=n p=1000 \cdot 0.001=1$.
The probability of exactly two elements failing:
$$
P_{1000}(2)=\frac{a^{2}}{2!} e^{-a}=\frac{1}{2 e} \approx 0.1831
$$
The probability of at l... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,076 |
Example 14. A telephone station serves 400 subscribers. For each subscriber, the probability that they will call the station within an hour is 0.01. Find the probabilities of the following events: "5 subscribers will call the station within an hour"; "no more than 4 subscribers will call the station within an hour"; "n... | Solution. According to the condition $n=400, p=0.01$, then $a=n p=$ $400 \cdot 0.01=4$.
In accordance with formula (3.3.1) we find
$$
\begin{gathered}
P_{400}(5)=\frac{4^{5}}{5!} e^{-4} \approx 0.1563 ; \\
P_{400}(0 \leq k \leq 4)=P_{400}(0)+P_{400}(1)+P_{400}(2)+P_{400}(3)+P_{400}(4) \approx \\
\approx 0.0183+0.0733... | 0.1563,0.6289,0.7619 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,077 |
Example 15. Independent trials are conducted, in each of which event $A$ may occur with a probability of $p=0.01$. Find the probability that in 100 trials, event $A$ will occur respectively $1,2,3,4,5,6$ times, and will not occur at all. | Solution. When calculating the desired probabilities, we will use the recurrence formula
$$
P_{n}(k+1)=\frac{a}{k+1} P_{n}(k)
$$
which is derived as follows:
$$
P_{n}(k+1)=\frac{a^{k+1}}{(k+1)!} e^{-a}=\frac{a}{k+1} \cdot \frac{a^{k}}{k!} e^{-a}=\frac{a}{k+1} P_{n}(k)
$$
When $a=1$, this formula becomes
$$
P_{n}(k... | \begin{aligned}P_{100}(0)&=0.3679,\\P_{100}(1)&=0.3679,\\P_{100}(2)&=0.1839,\\P_{100}(3)&=0.0613,\\P_{100}(4)&=0.0153,\\P_{} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,078 |
Example 16. At the faculty, 500 students are enrolled. What is the probability that September 1st is the birthday of $k$ students of this faculty? Calculate this probability for the values $k=0,1,2,3$. | Solution. Since $n=500$ is quite large and $p=1 / 365$ is quite small, we can assume that the random number $X$ of students born on September 1 follows a Poisson distribution with parameter $a=n p=500 / 365=1.36986$.
Using formula (3.3.1), we get $p_{0}=P(X=0)=e^{-a} \approx 0.2541$. Using formula (3.3.8), we sequenti... | p_{0}\approx0.2541,p_{1}\approx0.3481,p_{2}\approx0.2385,p_{3}\approx0.1089 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,079 |
Example 17. When the polio vaccine is administered, immunity is developed in $99.99\%$ of cases. What is the probability that out of 1000 vaccinated children, $1, 2, 3, 4$ children will get sick, respectively? | Solution. The probability of getting sick $p=0.0001$, the number of trials $n=10000$, so $a=np=1$. According to formula (3.3.1), we find
$$
P(1)=\frac{1}{1!} e^{-1}=0.3679
$$
In accordance with formula (3.3.8), we obtain:
$$
\begin{aligned}
& P(2)=\frac{1^{2}}{2!} e^{-1}=\frac{1}{2} P(1)=0.1839, \quad P(3)=\frac{e^{... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,080 |
Example 2. The random variable $X$ is uniformly distributed on the interval $[\alpha, \beta]$. Find the probability of its values falling into the interval $(\gamma, \delta)$, which belongs to the interval $[\alpha, \beta]$. | Solution. Using formula (2.3.3), we find
$$
\begin{aligned}
P(\gamma<X<\delta)= & \int_{\gamma}^{\delta} p(x) d x=\int_{\gamma}^{\delta} \frac{1}{\beta-\alpha} d x=\frac{1}{\beta-\alpha} \int_{\gamma}^{\delta} d x=\frac{\delta-\gamma}{\beta-\alpha} \\
& P(\gamma<X<\delta)=\frac{\delta-\gamma}{\beta-\alpha}
\end{aligne... | P(\gamma<X<\delta)=\frac{\delta-\gamma}{\beta-\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,082 |
## Example 5.
Find the mathematical expectation of a random variable $X, \quad$ having a uniform distribution on the interval $[\alpha, \beta]$. | Solution.
Using formula $(2.4 .8)$ and taking into account the formula, we find
$$
\begin{gathered}
M(x)=\int_{-\infty}^{+\infty} x p(x)=\int_{-\infty}^{\alpha} x p(x) d x+\int_{\alpha}^{\beta} x p(x) d x+\int_{\beta}^{+\infty} x p(x) d x= \\
=\int_{\alpha}^{\beta} \frac{x d x}{\beta-\alpha}=\left.\frac{1}{\beta-\alp... | \frac{\beta+\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,083 |
Example 6. Find the variance of a random variable $X$ that has a uniform distribution on the interval $[\alpha, \beta]$. | Solution. Using formula (2.5.15) and taking into account formulas (3.4.3) and (3.4.7), we find 204
$$
\begin{aligned}
D(X) & =\int_{-\infty}^{+\infty}(x-a)^{2} p(x) d x=\int_{-\infty}^{\alpha}(x-a)^{2} p(x) d x+\int_{\alpha}^{\beta}(x-a)^{2} p(x) d x+ \\
& +\int_{\beta}^{+\infty}(x-a)^{2} p(x) d x=\int_{\alpha}^{\beta... | D(X)=\frac{(\beta-\alpha)^{2}}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,084 |
Example 8. The random variable $X$ is uniformly distributed on the interval $[-3,2]$. Find the distribution function $F(x)$ of this random variable. | Solution. The distribution function $F(x)$ of a random variable $X$, uniformly distributed on the interval $[\alpha, \beta]$, is defined by formula (3.4.6). From the problem statement, it follows that $\alpha=-3, \beta=2, \beta-\alpha=2-(-3)=5$; therefore,
$$
F(x)=\left\{\begin{array}{cll}
0 & \text { for } & x \leq-3... | F(x)={\begin{pmatrix}0&\text{for}&x\leq-3\\\frac{x+3}{5}&\text{for}&-3<x<2\\1&\text{for}&x\geq2\end{pmatrix}.} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,085 |
Example 9. Find the mathematical expectation of a random variable $X$,
uniformly distributed on the interval $[2,8]$. | Solution. The mathematical expectation of a random variable $X$, uniformly distributed on the interval $[\alpha, \beta]$, is defined by formula (3.4.7). Since in this case $\alpha=2, \beta=8$, then
$$
M(X)=\frac{8+2}{2}=5
$$ | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,086 |
Example 1. Prove that the function (3.5.1), defining the density of the normal distribution, satisfies condition (2.3.6), i.e.
$$
\int_{-\infty}^{+\infty} p(x) d x=1
$$ | Solution. In the integral
$$
\int_{-\infty}^{+\infty} p(x) d x=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{+\infty} e^{-(x-a)^{2} / 2 \sigma^{2}} d x
$$
we will transition to a new variable \( t \) using the formula
$$
t=\frac{x-a}{\sigma}
$$
Then \( x=a+\sigma t, d x=\sigma d t \). Since the new limits of integr... | 1 | Calculus | proof | Yes | Yes | olympiads | false | 33,088 |
Example 3. Find the variance and standard deviation of a normal random variable. | Solution. Using formulas (2.5.15) and (3.5.11), we get
$$
D(X)=\int_{-\infty}^{+\infty}(x-a)^{2} p(x) d x=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{+\infty}(x-a)^{2} e^{-(x-a)^{2} / 2 \sigma^{2}} d x
$$
By transitioning to a new variable $t$ according to formula (3.5.9), we get
$$
D(X)=\frac{1}{\sigma \sqrt{2 \p... | D(X)=\sigma^{2},\sigma=\sqrt{D(X)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,089 |
Example 4. Prove that the Laplace function is odd. Solution. Indeed,
$$
\Phi(-x)=\frac{1}{\sqrt{2 \pi}} \int_{0}^{-x} e^{-t^{2} / 2} d t=-\frac{1}{\sqrt{2 \pi}} \int_{0}^{x} e^{-u^{2} / 2} d u=-\frac{1}{\sqrt{2 \pi}} \int_{0}^{x} e^{-t^{2} / 2} d t=-\Phi(x)
$$
Here, using the substitution $u=-t$ from the first integr... | Solution. Taking into account the definition of the improper integral and formula (3.5.3), we obtain
$$
\lim _{x \rightarrow+\infty} \Phi(x)=\lim _{x \rightarrow+\infty} \frac{1}{\sqrt{2 \pi}} \int_{0}^{x} e^{-t^{2} / 2} d t=\frac{1}{\sqrt{2 \pi}} \int_{0}^{+\infty} e^{-t^{2} / 2} d t=\frac{1}{\sqrt{2 \pi}} \cdot \fra... | \frac{1}{2} | Calculus | proof | Yes | Yes | olympiads | false | 33,090 |
Example 6. Find the distribution function $F(x)$ of a random variable $X$ that is distributed according to the normal law. | Solution. According to formula (2.3.2) and the definition of the normal distribution (3.5.1), we obtain
$$
F(x)=\int_{-\infty}^{x} \frac{1}{\sigma \sqrt{2 \pi}} e^{-(t-a)^{2} / 2 \sigma^{2}} d t=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{x} e^{-(t-a)^{2} / 2 \sigma^{2}} d t
$$
We switch to the new variable $u=\fra... | F(x)=\frac{1}{2}+\Phi(\frac{x-}{\sigma}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,091 |
Example 7. Calculate the central moments of a random variable distributed according to the normal law. | Solution. According to the definition (see formula (2.6.7)) and formula (3.5.1), we obtain
$$
\mu_{k}=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{+\infty}(x-\dot{a})^{k} e^{-(x-a)^{2} / 2 \sigma^{2}} d x
$$
We transition to a new variable $t$ using the formula $t=\frac{x-a}{\sigma}$, then $x=a+\sigma t, d x=\sigma ... | \mu_{2}=1\cdot3\ldots(2-1)\sigma^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,092 |
Example 9. Write the probability density function and the distribution function of the normal random variable $X$, if $M(X)=3, \quad D(X)=4$. | Solution. From the condition, it follows that $a=M(X)=3, \sigma(x)=\sqrt{D(X)}=$ $=2$. Substituting these values into formula (3.5.1), we find the density function of the given random variable $X$:
$$
p(x)=\frac{1}{2 \sqrt{2 \pi}} e^{-(x-3)^{2} / 2 \cdot 2^{2}}=\frac{1}{2 \sqrt{2 \pi}} e^{-(x-3)^{2} / 8}
$$
According... | p(x)=\frac{1}{2\sqrt{2\pi}}e^{-(x-3)^{2}/8},\quadF(x)=\frac{1}{2}+\Phi(\frac{x-3}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,093 |
Example 10. The random variable $X$ is distributed according to the normal law, with $M(X)=10, \quad D(X)=4$. Find:
$$
\text { 1) } P(12<X<14) ; 2) P(8<X<12)
$$ | Solution. From the condition, it follows that $a=10, \sigma(x)=\sqrt{D(X)}=2$. According to formula (3.5.2), we find:
$$
\begin{gathered}
P(12<X<14)=\Phi\left(\frac{14-10}{2}\right)-\Phi\left(\frac{12-10}{2}\right)=\Phi(2)-\Phi(1)= \\
=0.477250-0.341345=0.135905 \\
P(8<X<12)=\Phi\left(\frac{12-10}{2}\right)-\Phi\left(... | 0.1359050.682690 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,094 |
Example 12. The weight of the caught fish follows a normal distribution with parameters $a=375$ g, $\sigma=25$ g. Find the probability that the weight of one fish will be: a) from 300 to 425 g; b) no more than 450 g; c) more than 300 g. | Solution. We use formula (3.5.2), setting $a=375$ g, $\sigma=25$ g.
In the first case $\alpha=300, \beta=425$, therefore
$$
\begin{gathered}
P(300< X < 425) = \Phi\left(\frac{425-375}{25}\right) - \Phi\left(\frac{300-375}{25}\right) = \\
= \Phi(2) - \Phi(-3) = \Phi(2) + \Phi(3) = 0.97725 + 0.49865 = 0.97725 + 0.49865... | 0.9987 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,095 |
Example 13. When measuring a part, random errors occur that follow a normal distribution with the parameter $\sigma=10$ mm. Find the probability that the measurement is made with an error not exceeding $15 \mathrm{MM}$. | Solution. We will use formula (3.5.4) and the table of values of the Laplace function. In this case, $\delta=15, \sigma=10$, so
$$
P(|X-a|<15)=2 \Phi\left(\frac{15}{10}\right)=2 \Phi(1.5)=2 \cdot 0.433193=0.866386
$$ | 0.866386 | Other | math-word-problem | Yes | Yes | olympiads | false | 33,096 |
Example 14. A machine manufactures bearings, which are considered suitable if the deviation $X$ from the design size in absolute value does not exceed 0.77 mm. What is the most probable number of suitable bearings out of 100, if the random variable $X$ is normally distributed with the parameter $\sigma=0.4$ mm? | Solution. First, we find the probability of deviation using formula (3.5.4) with $\delta=0.77$ and $\sigma=0.4$:
$$
P(|X-a|<0.77)=2 \Phi\left(\frac{0.77}{0.4}\right) \approx 2 \Phi(1.93)=2 \cdot 0.473197=0.946394
$$
Approximating $p=0.95$ and $q=0.05$, according to formula (3.1.5), i.e., $n p-q \leq k_{0} \leq n p+p$... | 95 | Other | math-word-problem | Yes | Yes | olympiads | false | 33,097 |
Example 15. An automatic machine produces rollers, controlling their diameters $X$. Assuming that the random variable $X$ is normally distributed, with parameters $a=10$ mm, $\sigma=0.1$ mm, find the interval in which the diameters of the produced rollers will be contained with a probability of 0.9973. | Solution. We will use formula (3.5.4). In this case, it is known that $a=10, \sigma=0.1, 2 \Phi(\delta / \sigma)=0.9973$; it is required to determine $\delta$ and the interval $(a-\delta, a+\delta)$.
From the table of values of the Laplace function, we find that $\delta / \sigma=3$. This follows from the equality $\Ph... | 0.682690 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,098 |
Example 18. Independent random variables $X, Y, Z$ are normally distributed, with $M(X)=1, D(X)=2 ; M(Y)=-2, D(Y)=3$; $M(Z)=5, D(Z)=4$. Write the probability density function and the distribution function of their sum. | Solution. Since $M(X+Y+Z)=M(X)+M(Y)+M(Z)=$ $=1+(-2)+5=4, \quad D(X+Y+Z)=D(X)+D(Y)+D(Z)=2+3+4=9, \quad$ then the sum $X+Y+Z$ has a normal distribution with parameters $a=4$, $\sigma=3$. Therefore,
$$
p(x)=\frac{1}{3 \sqrt{2 \pi}} e^{-(x-4)^{2} / 18}, \quad F(x)=\frac{1}{2}+\Phi\left(\frac{x-4}{3}\right)
$$ | p(x)=\frac{1}{3\sqrt{2\pi}}e^{-(x-4)^{2}/18},\quadF(x)=\frac{1}{2}+\Phi(\frac{x-4}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,099 |
Example 19. Find the probability that a normal random variable with a mean of 3 and a variance of 4 will take values: 1) in the interval $(-1,5)$; 2) no more than $8$; 3) no less than 5; 4) in the interval $(-3,9)$. | Solution. We will use formula (3.5.2).
1) According to the condition, $\alpha=-1, \beta=5, a=3, \sigma=\sqrt{4}=2$, therefore
$$
\begin{gathered}
P(-1<X<5)=\Phi\left(\frac{5-3}{2}\right)-\Phi\left(\frac{-1-3}{2}\right)=\Phi(1)-\Phi(-2)=\Phi(1)+\Phi(2)= \\
=0.3413+0.4772=0.8185
\end{gathered}
$$
2) $P(X \leq 8)=P(-\i... | 0.8185,0.9938,0.1587,0.9972 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,100 |
Example 20. The diameter of a part manufactured in the workshop is a random variable distributed according to the normal law with parameters $a=4.5$ cm and $\sigma=0.05$ cm. Find the probability that the diameter size of a randomly taken part differs from the mathematical expectation by no more than $1$ mm. | Solution. Applying formula (3.5.4) with $\delta=0.1$ cm, we obtain the required probability.
$$
P(|X-4.5| \leq 0.1)=2 \Phi\left(\frac{0.1}{0.05}\right)=2 \Phi(2)=2 \cdot 0.4772=0.9544
$$ | 0.9544 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,101 |
Example 21. The random variable $X$ is distributed according to the normal law. The mathematical expectation and variance of this variable are 7 and 16, respectively. Find the probability that the deviation of the variable $X$ from its mathematical expectation in absolute value does not exceed two. | Solution. According to the condition $M(X)=7, D(X)=16, \sigma=\sqrt{16}=4, \delta=2$. In accordance with formula (3.5.4) we find
$$
P(|X-7| \leq 2)=2 \Phi\left(\frac{2}{4}\right)=2 \cdot \Phi(0.5)=2 \cdot 0.1915=0.3830
$$ | 0.3830 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,102 |
Example 22. The root mean square deviation of a random variable distributed according to the normal law is $2 \mathrm{~cm}$, and the mathematical expectation is 16 cm. Find the limits within which with a probability of 0.95 the value of the random variable should be expected. | Solution. According to formula (3.5.4) when $\sigma=2$, we have $2 \Phi\left(\frac{\delta}{\sigma}\right)=0.95$ or $\Phi\left(\frac{\delta}{\sigma}\right)=0.475$. From the table of values of the Laplace function, we find that $\delta / 2=1.96$. Therefore, $\delta / 2=1.96$, from which $\delta=3.92$.
Thus, with a proba... | 12.08\leqX\leq19.92 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,103 |
Example 23. The random variable $X$ is distributed according to the normal law. The mathematical expectation $a=0$ and the standard deviation of this variable $\sigma=0.5$. Find the probability that the deviation of the random variable $X$ in absolute value will be less than one. | Solution. We will use formula (3.5.4), which when $a=0$ takes the form
$$
P(|X|<\delta)=2 \Phi\left(\frac{\delta}{\sigma}\right)
$$
Since according to the condition $a=0, \sigma=0.5, \delta=1$, then
$$
P(|X|<1)=2 \Phi\left(\frac{1}{0.5}\right)=2 \Phi(2)=2 \cdot 0.4772=0.9544
$$ | 0.9544 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,104 |
Example 24. When manufacturing a certain product, its weight $X$ is subject to random fluctuations. The standard weight of the product is 30 g, its standard deviation is 0.7, and the random variable $X$ is normally distributed. Find the probability that the weight of a randomly selected product is within the range of 2... | Solution. We will use formula (3.5.2): In this case, $a=30$, $\sigma=0.7$, $\alpha=28$, $\beta=31$. According to the specified formula, we find
$$
\begin{gathered}
P(28<X<31)=\Phi\left(\frac{31-30}{0.7}\right)-\Phi\left(\frac{28-30}{0.7}\right)=\Phi(1.43)-\Phi(-2.86)= \\
=\Phi(1.43)+\Phi(2.86)=0.424+0.498=0.922
\end{g... | 0.922 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,105 |
Example 3. Find the variance of a random variable $X$ that has a geometric distribution. | Solution. To calculate the variance, we will use formula $(2.5 .4)$ and the formulas for the sums of series:
$$
\sum_{m=1}^{\infty} m q^{m-1}=\frac{1}{(1-q)^{2}}, \sum_{m=2}^{\infty} m(m-1) q_{-}^{m-2}=\frac{2}{(1-q)^{3}}
$$
First, we find the expected value of the square of the quantity $X$:
$$
\begin{aligned}
& M\... | \frac{1-p}{p^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,107 |
Example 4. Prove that the function (3.6.2) satisfies condition (2.3.6), i.e. $\int_{-\infty}^{+\infty} p(x) d x=1$. | Solution. Indeed,
$$
\int_{-\infty}^{+\infty} p(x) d x=\int_{-\infty}^{0} p(x) d x+\int_{0}^{+\infty} p(x) d x=\int_{-\infty}^{0} 0 \cdot d x+\int_{0}^{+\infty} \alpha e^{-\alpha x} d x=
$$
$$
=-\int_{0}^{+\infty} d\left(e^{-\alpha x}\right)=-\left.e^{-\alpha x}\right|_{0} ^{+\infty}=-\left(e^{-\infty}-e^{0}\right)=1... | 1 | Calculus | proof | Yes | Yes | olympiads | false | 33,108 |
Example 5. Find the distribution function of a random variable distributed according to the exponential law. | Solution. We will use formula (2.3.2). For $x \leq 0$
$$
F(x)=\int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{x} 0 d t=0
$$
If $x>0$, then
$$
\begin{aligned}
& F(x)=\int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{0} p(t) d t+\int_{0}^{x} p(t) d t=\int_{-\infty}^{0} 0 d t+ \\
& +\int_{0}^{x} \alpha e^{-\alpha t} d t=-\left.e... | F(x)=\begin{cases}0&\text{for}x<0\\1-e^{-\alphax}&\text{for}x\geq0\end{cases} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,109 |
Example 6. Find the mathematical expectation of a random variable $X$, having an exponential distribution. | Solution. We will use formula (2.4.8) and formula (3.6.2):
$$
M(X)=\int_{-\infty}^{+\infty} x p(x) d x=\int_{-\infty}^{0} x p(x) d x+\int_{0}^{+\infty} x p(x) d x=\int_{-\infty}^{0} 0 \cdot x d x+\int_{0}^{+\infty} x \alpha e^{-\alpha x} d x
$$
$$
M(X)=\alpha \int_{0}^{+\infty} x e^{-\alpha x} d x
$$
Applying the in... | M(X)=\frac{1}{\alpha},\sigma(x)=\frac{1}{\alpha} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,110 |
Example 8. Find the probability of falling into the interval $(a, b)$ of values of a random variable $X$, distributed according to the exponential law. | Solution. We will use the distribution function (3.6.7) and formula $(2.2 .10)$ :
$$
P(a0$ and $F(x)=0$ for $x \leq 0, F(b)=1-e^{-\alpha b}$, $F(a)=1-e^{-\alpha a}$, ro
$$
\begin{gathered}
P(a<X<b)=F(b)-F(a)=\left(1-e^{-\alpha b}\right)-\left(1-e^{-\alpha a}\right) \\
P(a<X<b)=e^{-\alpha a}-e^{-\alpha b}
\end{gathere... | e^{-\alpha}-e^{-\alphab} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,111 |
Example 11. Find the mathematical expectation of the random variable
$X$, the density of which is determined by the function $p(x)=0.2 e^{-0.2 x}$ for $x \geq 0$. | Solution. Since in this case $\alpha=0.2$ and $M(X)=\frac{1}{\alpha}$, then
$$
M(X)=\frac{1}{2 / 10}=\frac{10}{2}=5, \quad M(X)=5
$$ | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,113 |
Example 14. In a batch of 12 parts, 8 are standard. Find the probability that among 5 randomly selected parts, 3 will be standard. | Solution. We apply formula (3.6.3). Since in this case $N=12, M=8, n=5, m=3$, then
$$
\begin{gathered}
P(X=3)=\frac{C_{M}^{m} C_{N-M}^{n-m}}{C_{N}^{n}}=\frac{C_{8}^{3} \cdot C_{4}^{2}}{C_{12}^{5}}=\frac{8!}{3!5!} \cdot \frac{4!}{2!2!}: \frac{12!}{5!7!}= \\
=\frac{8!4!5!7!}{3!5!2!2!12!}=\frac{8!4!7!}{3!2!2!12!}=\frac{7... | \frac{14}{33} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,114 |
Example 3. Evaluate the probability that in 3600 independent throws of a die, the number of times 6 points appear will be no less than 900. | Solution. Let $X$ be the number of occurrences of 6 points in 3600 tosses, then $M(X)=3600 \cdot \frac{1}{6}=600$. Using inequality (4.1.2.) with $a=600, \delta=900$:
$$
P(X \geq 900) \leq \frac{600}{900}=\frac{2}{3}
$$ | \frac{2}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,116 |
Example 4. The random variable $X$ has a variance $D(X)=0.001$. What is the probability that the random variable $X$ differs from $M(X)=a$ by more than 0.1? | Solution. By the first Chebyshev's inequality
$$
P(|X-a| \geq 0.1) \leq \frac{D(X)}{(0.1)^{2}}=\frac{0.001}{0.01}=0.1
$$ | 0.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,117 |
Example 5. The random variable $X$ has a variance $D(X)=0.004$. Find the probability that the random variable $X$ differs from $M(X)$ by more than 0.2. | Solution. According to inequality (4.1.3), we get
$$
P(|X-a|>0.2)<\frac{0.004}{(0.2)^{2}}=\frac{0.004}{0.04}=0.1
$$ | 0.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,118 |
Example 8. The average length of a part is $50 \mathrm{~cm}$, and the variance is 0.1. Estimate the probability that the length of the manufactured part will be no less than 49.5 cm and no more than 50.5 cm. | Solution. Since here $a=50$, the condition $49.5<X<50.5$, in which the random variable $X$ denotes the possible length of the part, is transformed by subtracting the number $a=50$ from each term to the form $|X-a| \leq 0.5$.
Applying inequality (4.1.4.) in the case $\varepsilon=0.5$ and $D(X)=0.1$, we get:
$$
P(|X-a|... | 0.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,119 |
Example 9. The germination rate of seeds of a certain crop is 0.75. Estimate the probability that out of 1000 sown seeds, the number of seeds that will germinate will be from 700 to 800 inclusive. | Solution. In this case, $M(X)=a=1000 \cdot 0.75=750$, the boundary values of the random variable $X$ are symmetric relative to $M(X)=750$. Therefore, from the original inequalities $700 \leq X \leq 800$, we can transition to the inequalities $-50 \leq X-750 \leq 50$, or $|X-a| \leq 50$, which gives the left part of ine... | 0.925 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,120 |
Example 10. The probability of producing a non-standard part under certain technological conditions is 0.1. Estimate the probability that the number of non-standard parts among 10000 will be within the range of 950 to 1030 inclusive. | Solution. The number of non-standard parts under the given conditions is a random variable $X$ distributed according to the binomial law. According to formulas (3.2.5.) and (3.2.6.), for $n=10000, p=0.1$, $q=0.9$, we get $M(X)=10000 \cdot 0.1=1000, D(X)=10000 \cdot 0.1 \cdot 0.9=900$. From this, it is clear that the bo... | 0.64 | Other | math-word-problem | Yes | Yes | olympiads | false | 33,121 |
Example 1. Find the probability that the frequency of getting a six in 10000 independent tosses of a die deviates from the probability of getting a six by less than 0.01 in absolute value. | Solution. We will use inequality (4.2.5). In this case, $n=10000, p=1 / 6, q=5 / 6$, so
$$
P\left(\left|\frac{m}{n}-p\right| \leq 0.01\right)=P\left(\left|\frac{m}{10000}-\frac{1}{6}\right|<0.01\right) \geq 1-\frac{\frac{1}{6} \cdot \frac{5}{6}}{10000 \cdot 0.01^{2}} \approx 0.86
$$ | 0.86 | Other | math-word-problem | Yes | Yes | olympiads | false | 33,122 |
Example 2. In the stamping of plastic plates, the defect rate is $3 \%$. Find the probability that when checking a batch of 1000 plates, the deviation from the established defect rate will be less than $1 \%$. | Solution. From the condition of the problem, it follows that $n=1000, \varepsilon=0.01$, $p=0.03, q=1-p=0.97$.
In accordance with formula (4.2.5.), we obtain
$$
P\left(\left|\frac{m}{n}-p\right| \leq 0.01\right) \geq 1-\frac{p q}{n \varepsilon^{2}}=1-\frac{0.03 \cdot 0.97}{10000 \cdot(0.01)^{2}}=1-\frac{0.0291}{0.1}=... | 0.709 | Other | math-word-problem | Yes | Yes | olympiads | false | 33,123 |
Example 3. For what number of independent trials will the probability of the inequality $\left|\frac{m}{n}-p\right|<0.2$ being satisfied exceed 0.96, if the probability of the event occurring in a single trial $p=0.7$? | Solution. According to the problem, we have: $\varepsilon=0.2, p=0.7$, so $q=0.3$; it is required to determine $n$ using inequality (4.2.5). The condition $P>0.96$ is equivalent to the inequality
$$
\frac{p q}{n \varepsilon^{2}} < 0.04
$$
or
$$
n > \frac{p q}{\varepsilon^{2} \cdot 0.04}
$$
Substituting the values $... | 132 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 33,124 |
Example 6. Determine how many measurements of the cross-sectional diameter of trees need to be made on a large plot so that the average diameter of the trees differs from the true value $a$ by no more than 2 cm with a probability of at least 0.95. The standard deviation of the cross-sectional diameter of the trees does... | Solution. We will consider the selection of trees for measurements such that the measurement results can be regarded as independent random variables. Let $X_{i}$ denote the measurement result of the cross-section of the $i$-th tree. According to the problem, $\sigma\left(X_{i}\right)=\sqrt{D\left(X_{i}\right)} \leq 10$... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,125 |
Example 8. Given a sequence of independent random variables $X_{1}, X_{2}, \ldots, X_{n}, \ldots$ each $X_{k}$ of which can take values: $-k \alpha, 0, k \alpha(\alpha=$ const $)$ with probabilities: $\frac{1}{2 k^{2}}, 1-\frac{1}{k^{2}}, \frac{1}{2 k^{2}}$. Can Chebyshev's theorem be applied to these variables? | Solution. To answer the given question, we will check the boundedness of the variances of these random variables by a single constant $C$ (the other conditions of Chebyshev's theorem are satisfied: the independence of the random variables and a sufficiently large number of them). For this, we will first find their math... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,126 |
Example 9. It is known that the variance of each of the given independent random variables does not exceed 4. Determine the number of such variables for which the probability of the deviation of the arithmetic mean of the random variable from the arithmetic mean of their mathematical expectations by no more than 0.25 e... | Solution. Inequality (4.2.1) for $C=4$ and $\varepsilon=0.25$ takes the form
$$
m\left(\left\lvert\, \frac{1}{n} \sum_{i=1}^{n} X_{i}-\frac{1}{n} \sum_{i=1}^{n} M\left(X_{i} \mid < 0.25\right) \geq 1-\frac{4}{n \cdot (0.25)^{2}}\right.\right.
$$
From the condition, it follows that
$$
\begin{aligned}
& 1-\frac{4}{n \... | n\geq6400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,127 |
Example 10. Starting from which number $n$ of independent trials is the inequality $p\left(\left|\frac{m}{n}-p\right|<0.1\right)>0.97$ satisfied, if in a single trial $p=0.8?$ | Solution. Inequality (4.2.5) in this case takes the form
$$
P\left(\left|\frac{m}{n}-p\right|<0.01\right)>0.97
$$
From this we find
$$
1-0.97>\frac{0.16}{0.01 \cdot n}, 0.03>\frac{0.16}{0.01 \cdot n}, n>\frac{0.16}{0.03 \cdot 0.01}=\frac{0.16}{0.0003} \approx 533.33
$$
Thus, $n \geq 534$.
## Problems
1. The proba... | n\geq534 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 33,128 |
Example 1. The probability of event $A$ occurring in each of 900 independent trials is $p=0.8$. Find the probability that event $A$ will occur: a) 750 times; b) 710 times. | Solution. From the condition, it follows that $n=900, p=0.8$, therefore $q=0.2$; in the first case $k=750$, in the second case $k=710$. Since $n p q=900 \cdot 0.8 \cdot 0.2=144>10$, we can use formula (4.3.1) or (4.3.2)-(4.3.4).
Consider the first case. Using formula (4.3.3), we determine $x$:
$$
x=\frac{750-900 \cdo... | 0.0236 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,129 |
Example 5. The germination rate of seeds of this plant is 0.9. Find the probability that out of 900 sown seeds, the number of sprouted seeds is between 790 and 830. | Solution. To find the required probability, we will use formula (4.3.7). From the condition, it follows that $n=900, k_{1}=790, k_{2}=830$, $p=0.9, q=0.1$. Using formulas (4.3.6), we determine
$$
\begin{gathered}
x_{1}=\frac{790-900 \cdot 0.9}{\sqrt{900 \cdot 0.9 \cdot 0.1}}=\frac{790-810}{\sqrt{81}}=-\frac{20}{9} \ap... | 0.9736 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,131 |
Example 6. Production gives $1 \%$ of defects. What is the probability that out of 1100 items taken for research, no more than $17$ will be defective? | Solution. From the condition, it follows that $n=1100, p=0.01, q=0.99$, $0 \leq k \leq 17$. Using formulas (4.3.6), we determine
$$
\begin{gathered}
x_{1}=\frac{0-1100 \cdot 0.01}{\sqrt{1100 \cdot 0.01 \cdot 0.99}}=\frac{-11}{\sqrt{10.89}}=-\frac{11}{3.3} \approx-3.33 \\
x_{2}=\frac{17-11}{3.3}=\frac{6}{3.3} \approx 1... | 0.9651 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,132 |
Example 7. The probability of manufacturing a first-class part on a given machine is 0.8. Find the probability that among 100 randomly taken parts, 75 will be of the first class. | Solution. We will find the required probability using formula (4.3.2). According to the condition, $n=100, k=75, p=0.8$; hence $q=1-p=0.2$. According to formula (4.3.3), we calculate
$$
x=\frac{k-n p}{\sqrt{n p q}}=\frac{75-100 \cdot 0.8}{\sqrt{100 \cdot 0.8 \cdot 0.2}}=\frac{75-80}{\sqrt{16}}=\frac{-5}{4}=-1.25
$$
f... | 0.04565 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,133 |
Example 8. The probability of an event occurring in each of 100 independent trials is constant and equal to $p=0.8$. Find the probability that the event will occur: a) not less than 70 and not more than 85 times; b) not less than 70 times; c) not more than 69 times. | Solution. We will use formula (4.3.7) in cases a) and b).
a) According to the condition, $n=100, p=0.8, q=0.2, k_{1}=70, k_{2}=85$. Using formulas (4.3.6), we calculate $x_{1}$ and $x_{2}$:
$$
x_{1}=\frac{70-100 \cdot 0.8}{\sqrt{100 \cdot 0.8 \cdot 0.2}}=\frac{70-80}{\sqrt{16}}=-\frac{10}{4}=-2.5
$$
$$
x_{2}=\frac{8... | 0.8882,0.9938,0.0062 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,134 |
Example 9. The probability of an event occurring in each of 900 independent trials is 0.5. Find the probability that the relative frequency of the event's occurrence deviates from its probability in absolute value by no more than 0.02. | Solution. We will use formula (4.3.9). According to the condition, $n=900, p=0.5, q=0.5, \varepsilon=0.02$. In accordance with the specified formula, we obtain
$$
P\left(\left|\frac{m}{900}-0.5\right| \leq 0.02\right)=2 \Phi\left(0.02 \sqrt{\frac{900}{0.5 \cdot 0.5}}\right)=2 \Phi(1.2)
$$
From the table, we find $\Ph... | 0.7698 | Other | math-word-problem | Yes | Yes | olympiads | false | 33,135 |
Example 10. 600 corn seeds are planted with a probability of 0.9 for each seed to germinate. Find the limit of the modulus of deviation of the frequency of germinated seeds from the probability $p=0.9$, if this limit is to be guaranteed with a probability $P=0.995$. | Solution. We will use formula (4.3.9), i.e.
$$
P\left(\left|\frac{m}{n}-p\right| \leq \varepsilon\right)=2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
$$
In this case, $n=600, p=0.9, q=1-0.9=0.1, P=0.995$; we need to determine the value of $\varepsilon$. The specified formula takes the form:
$$
\begin{gathered... | 0.0034 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,136 |
Example 11. On average, $85 \%$ of the items coming off the conveyor are of the first grade. How many items need to be taken so that with a probability of 0.997, the deviation of the frequency of first-grade items from the probability $p=0.85$ in absolute value does not exceed $0.01?$ | Solution. We will use the formula
$$
P\left(\left|\frac{m}{n}-p\right| \leq \varepsilon\right)=2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
$$
From the condition, it follows that $p=0.85, q=1-0.85=0.15, \varepsilon=0.01$, $P=0.997$. We need to determine the value of $n$. Since
$$
2 \Phi\left(\varepsilon \sqrt... | 11171 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,137 |
Example 13. The probability of an event occurring in each of 900 independent trials is 0.5. Find such a positive number $\varepsilon$, so that with a probability of 0.77, the absolute deviation of the frequency of the event's occurrence from its probability 0.5 does not exceed $\varepsilon$. | Solution. We will use formula (4.3.9). In this case, $n=900, p=0.5, q=1-0.5=0.5$. Therefore,
$$
\begin{gathered}
2 \Phi\left(\varepsilon \sqrt{\frac{900}{0.5 \cdot 0.5}}\right)=0.77, \quad \text { or } \\
2 \Phi(60 \varepsilon)=0.77, \Phi(60 \varepsilon)=0.385
\end{gathered}
$$
From the table of values of the Laplace... | 0.02 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,138 |
Example 14. The quality control department checks 475 items for defects. The probability that an item is defective is 0.05. Find, with a probability of 0.95, the boundaries within which the number $m$ of defective items among those checked will be enclosed. | Solution. We will use formula (4.3.9). First, we determine the number $\varepsilon>0$, and then the boundaries within which the number $m$ is confined. According to the condition $n=475, p=0.05, q=0.95, P=0.95$. In accordance with the condition
$$
2 \Phi\left(\varepsilon \sqrt{\frac{475}{0.05 \cdot 0.95}}\right)=0.95,... | 15\leq\leq33 | Other | math-word-problem | Yes | Yes | olympiads | false | 33,139 |
Example 15. A die is rolled 80 times. Find, with a probability of 0.99, the boundaries within which the number $m$ of fives will be contained. | Solution. We will use formula (4.3.9). First, we will find the number $\varepsilon>0$, and then the boundaries within which the number $m$ is confined. According to the condition $n=80$; in this case $p=1 / 6, q=5 / 6$. In accordance with the condition
$$
\begin{gathered}
2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\r... | 5\leq\leq22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,140 |
Example 16. 500 items of the product manufactured by the enterprise are inspected, where the defect rate is $2 \%$. Find the probabilities that: 1) among them there will be exactly 10 defective items; 2) the number of defective items is between 10 and 20. | Solution. From the condition, it follows that $n=500, p=0.02$, therefore $q=1-p=0.98, n p=500 \cdot 0.02=10, n p q=500 \cdot 0.02 \cdot 0.98=9.8$.
According to formula (4.3.2), we find
$$
P_{500}(10)=\frac{1}{\sqrt{9.8}} \varphi\left(\frac{10-10}{\sqrt{9.8}}\right)=\frac{1}{\sqrt{9.8}} \varphi(0)=0.319 \cdot 0.399=0.... | 0.1270.499 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,141 |
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