# 1. Problem 11

## 1.1. Question

There are $$841$$ pupils in some school. Probability, that randomly chosen pupil has at least one D is equal to $$\frac{1}{12}$$. Calculate probability $$p$$, that number of pupils that have at least one D differs from $$\frac{841}{12}$$ at most by 20. Consider absolute difference!

1. $$p \geq 0.99371$$
2. $$p \approx 0.98741$$
3. $$p \approx 0.78741$$
4. $$p < 0.89371$$
5. $$p \leq 0.88741$$

## 1.2. Answer

2 is true; the rest is false; 5 is impossible

If $$P(X=k)$$ denotes the probability that exactly k pupils have at least one D, then $$P(X=k) = \binom{841}{k}(\frac{1}{12})^k (\frac{11}{12})^{841-k}$$

$$P(|X-\frac{841}{12}|<20) = P(X \in \{51,52,\ldots,89,90\})$$

# 2. Problem 12

## 2.1. Question

Segment $$DG$$ of the length $$1$$ is divided at random at two points. Probability that the lengths of three subsequent (counting from point D to G) segments form a decreasing sequence, is:

1. $$\frac{1}{2}$$
2. $$\frac{2}{3}$$
3. $$\frac{1}{4}$$
4. $$\frac{1}{3}$$
5. $$\frac{1}{6}$$

## 2.2. Answer

5 is true; the rest is false

# 3. Problem 13

## 3.1. Question

We toss a symmetric coin. What's the minimal number of tosses assuring with probability not less than $$0.88$$ that the absolute value of difference between the frequency of heads and $$\frac{1}{2}$$ is less than $$0.04$$?

1. 216
2. 126
3. 378
4. 252
5. 567

## 3.2. Answer

3 is correct; the rest is wrong

Derivation:

$$X_i \in \{0,1\}$$ - result of i-th cion toss; $$\mu = \frac{1}{2}$$; $$\sigma^2 = \frac{1}{4}$$

$$P(|\frac{1}{n} \sum_{i=1}^n (X_i - \frac{1}{2})| < q) \geq p$$

$$|\frac{1}{n} \sum_{i=1}^n (X_i - \frac{1}{2})| = \frac{\sigma}{\sqrt{n}} |\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^n (X_i - \mu)|$$

Therefore we get:

$$P( |\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^n (X_i - \mu)| < \frac{\sqrt{n}}{\sigma} q) \geq p$$

From the **Central limit theorem** we know that our distribution can be approximated by $$N(1,0)$$. Therefore:

$$2 \Phi ( \frac{\sqrt{n}}{\sigma} q) \geq p$$

And the general formula is:

$$2 \Phi ( 2 q \sqrt{n}) = p$$

For our data $$p = 0.88$$, $$q = 0.04$$ we get $$n = \lceil 377.7 \rceil = 378$$

# 4. Problem 14

## 4.1. Question

There are 5 ordinary dies and 2 erroneous having 6 on all sides. A randomly chosen die was rolled 4 times, and all 6 were observed. What's the probability that the die was ordinary?

1. $$\frac{6}{2597}$$
2. $$\frac{5}{2597}$$
3. $$\frac{1}{2597}$$
4. $$\frac{2}{2597}$$
5. $$\frac{4}{2597}$$

## 4.2. Answer

2 is correct; the rest is wrong

It can be solved using Bayes' theorem.

# 5. Problem 15

## 5.1. Question

We calculate the integral $$I = \int_0^1 \sqrt{(sin \pi x)} dx$$ using Monte-Carlo method. To do this we generate sequence $$\{X_i\}_{i \geq 1}$$ of independent random variables uniformly distributed on (0;1). This integral is approximated by the quantity $$A(n) = \frac{1}{n} \sum_{i=1}^n \sqrt{sin (\pi X_i)}$$. Let us notice that $$Var(\sqrt{sin(\pi X)}) \leq \frac{1}{\pi} - \frac{1}{\pi^2} = 0.21699$$, $$\sqrt{0.217}=0.4658.$$. 8100 simulations were performed. Absolute error $$e$$ of approximation of $$I$$ by $$A(n)$$ is with probability $$0.94$$:

1. $$e \geq 1.9468 \times 10^{-2}$$
2. $$e \geq 9.7342 \times 10 ^ {-3}$$
3. $$e < 6.4894 \times 10^{-3}$$
4. $$e = 8.0468 \times 10^{-3}$$
5. $$e > 1.1681 \times 10^{-2}$$

## 5.2. Answer

2 is true; the rest is false; 1 is impossible

Problem can be solved using the Central limit theorem.

# 6. Problem 16

## 6.1. Question

Random variables $$(X,Y)$$ have uniform distribution on set $$D=(x,y): y \in [0,1] \wedge x \in [y-1,y]$$. Error $$e$$ of approximation of $$X$$ by the best (in the mean square sense) Borel function $$Y$$ is:

1. $$e \in (\frac{1}{8},\frac{1}{3}]$$
2. $$e < \frac{1}{9}$$
3. $$e \in [\frac{1}{12},\frac{1}{8}]$$
4. $$e > \frac{1}{5}$$

## 6.2. Answer

3 is true; the rest is false

$$e = V(X)(1 - \rho_{X,Y}^2)$$

$$E(X) = 0$$, $$E(Y)=\frac{1}{2}$$, $$E(X^2) = \frac{1}{6}$$, $$E(Y^2)=\frac{1}{3}$$, $$E(XY) = \frac{1}{12}$$

$$e = \frac{1}{12}$$

# 7. Problem 17

## 7.1. Question

Random variable $$(X,Y)$$ has uniform distribution on set $$D = \{(x,y): |x| \leq 6; 0 \leq y \leq 6-|x|\}$$. Knowing that $$V(X)=6$$, $$V(Y) = 2$$, $$E(X^4) = \frac{432}{5}$$, $$E(X^2 Y) = \frac{36}{5}$$ find nonlinear regression of Y on X.

1. $$|X|/2$$
2. $$0$$
3. $$\frac{36}{3}-\frac{X^2}{3}$$
4. $$6-X$$
5. $$\frac{6}{2} - \frac{|X|}{2}$$

## 7.2. Answer

5 is true; the rest is false (probably)

It can be solved by looking at the plot.

# 8. Problem 18

## 8.1. Question

Let characteristic X have distribution with the density:

$$f_p(x) = \begin{cases} \frac{p}{2|x|^{p+1}} & \text{for |x| \geq 1} \\ 0 & \text{for |x| < 1} \end{cases}$$

where $$p>3$$.

It is known that: $$\int_1^\infty x f_p(x)dx = \frac{p}{2(p-1)}$$, $$\int_1^\infty x^2 f_p(x)dx = \frac{p}{2(p-2)}$$, $$\int_1^\infty ln(x) f_p(x) dx = \frac{1}{2p}$$.

Let us denote: $$\overline{|X|}_n=\frac{1}{n} \sum_{i=1}^n |X_i|$$, $$\overline{ln|X|}_n = \frac{1}{n} \sum_{i=1}^n ln|X_i|$$, $$\overline{X^2}_n = \frac{1}{n} \sum_{i=1}^n X_i^2$$.

Statistic $$1+\frac{1}{\overline{X}_n-1}$$

1. is an unbiased estimator of the parameter $$p/2$$
2. is an estimator of $$p$$ having minimal variance
3. is a consistent estimator of the parameter $$p$$
4. is not a consistent estimator of the parameter $$p$$
5. is an unbiased estimator of the parameter $$p$$

## 8.2. Answer

3 is true; the rest is false

# 9. Problem 19

## 9.1. Question

On the probability space $$(<-3,3>, B(<-3,3>),\frac{1}{6}|.|)$$ ( $$|.|$$ - Lebesgue measure) we define a random variable:

$$X(\omega) = \begin{cases} 1 & \text{for \omega \in [-3;-\frac{3}{2})} \\ -2\omega & \text{for \omega \in [-\frac{3}{2};\frac{3}{2}} \\ -2+2\omega & \text{for \omega \in [\frac{3}{2};3]} \end{cases}$$

1. mixed
2. uniform
3. of a continuous type
4. of discrete type
5. singular

## 9.2. Answer

1 is true; the rest is false; 2 is impossible (is it?)

# 10. Problem 20

## 10.1. Question

Let $$\{X_i\}, i \geq 1$$ be a sequence of uncorrelated random variables, such that $$E(X_i)=4^{-i}; i=1,2,\cdots$$, $$Var(X_i) = 2 - 5^{-i}$$. $$\frac{1}{n} \sum_{i=1}^n X_i$$ converges in probability to:

1. does not converge
2. 2
3. 0
4. 1
5. -1

## 10.2. Answer

3 is true; the rest is false

$$\frac{1}{n} \sum_{i=1}^n X_i \rightarrow 0$$

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