# 1. Problem 1

## 1.1. Question

Random variable $$X$$ have uniform distribution on $$[0;3]$$. Let $$Z = \max(X,0)$$, $$W = \min (X,0)$$. Is it true:

1. $$P(Z>=0) > P(W>=0)$$
2. $$P(Z>0) = P(W>0)$$
3. $$P(Z>0) = P(W<0)$$
4. $$P(Z<0) = P(W<0)$$

4 is true; the rest is false

# 2. Problem 2

## 2.1. Question:

Random variable $$Y$$ has uniform distribution on the interval $$(-4,4)$$. Let $$Y' = sign(Y)$$ and $$S_1 = |Y|$$. $$P(0 < Y'S_1 < 1/2)$$ is equal:

1. $$\frac{1}{16}$$
2. $$1/3$$
3. $$3/4$$
4. $$3/16$$

1 is true; the rest is false

# 3. Problem 3

## 3.1. Question

Random variable $$X$$ has uniform distribution on the interval $$(0,4)$$ and $$Y$$ is independent with two point distribution $$P(Y= -1) = 1 - P(Y=1) = 1/3$$.

1. $$V(X-Y) \in (0.5; 2.1222]$$
2. $$V(X-Y) > V(X+Y)$$
3. $$V(X-Y) >= \frac{29}{9}$$
4. $$V(X-Y) <= \frac{20}{9}$$
5. $$V(X-Y) >= 3.7222$$

4 is true; the rest is false

The exact result: $$V(X-Y) = \frac{20}{9}$$.

The way to obtain it:

$$V(X-Y) = E((X-Y)^2) - (E(X-Y))^2 = E(X^2)-2E(XY)+E(Y^2)-(E(X)-E(Y))^2$$

$$E(X)=2; E(Y)=\frac{1}{3}; E(X^2)=\frac{16}{3}; E(Y^2)=1; E(XY)=\frac{2}{3}$$

We simply substitute values and get the exact result.

# 4. Problem 4

## 4.1. Question

Let $$E$$ and $$K$$ be independent events. Then:

1. $$E$$ and $$K$$ exclude themselves.
2. $$E^C$$ and $$K^C$$ exclude themselves.
3. $$E$$ and $$K^C$$ can be dependent.
4. $$E \cap K$$ is an impossible event.
5. $$E^C$$ and $$K^C$$ are independent.

5 is true; the rest is false

# 5. Problem 5

## 5.1. Question

Let $$E$$ and $$K$$ be any random events. Assume that $$0 < P(E),P(K)<1$$. Which of the statements is true:

1. $$P(E \cap K) < P(E|K)P(K)$$
2. $$P(E \cup K) > P(E)+P(K)$$
3. $$P(K \cap E) > min(P(E), P(K))$$
4. $$P(E \cup K) < max(P(E),P(K))$$
5. $$P(E^C)+P(K^C)+P(E \cup K) - P(E^C \cup K^C) = 1$$

5 is true; the rest is false

# 6. Problem 6

## 6.1. Question

Random variables $$X$$ and $$Y$$ are independent and have the same uniform distributions on the interval $$(0,3)$$.

1. $$P(X-Y>-1) > \frac{7}{9}$$
2. $$P(X-Y>-1) < \frac{2}{9}$$
3. $$P(X-Y>-1) \in (\frac{2}{9},\frac{1}{2}]$$
4. $$P(X-Y>-1) = \frac{23}{27}$$
5. $$P(X-Y>-1) \in (\frac{1}{2},\frac{7}{9}]$$

5 is true; the rest is false

The answer can be read from a simple drawing without any calculations.

# 7. Problem 7

## 7.1. Question

Density of a random variable X is equal to:

$$f_X(x) = \begin{cases} \frac{1}{8} & \text{for -1 <= x <0} \\ \frac{3}{8} & \text{for 0<=x<2} \\ \frac{1}{16} & \text{for 2 <= x <= 4} \end{cases}$$

1. $$P(-\frac{1}{2} < X < 3) = \frac{7}{8}$$
2. $$P(-\frac{1}{2} < X < 3) \leq \frac{1}{2}$$
3. $$P(-\frac{1}{2} < X < 3) = \frac{13}{16}$$
4. $$P(-\frac{1}{2} < X < 3) < \frac{3}{8}$$
5. $$P(-\frac{1}{2} < X < 3) \geq \frac{15}{16}$$

1 is correct; the rest is wrong

Moreover please note that by simple logic answer 4 cannot be true, because if it was, then answer 2 would be true too, and as it was written in the introduction to these problems, there's is only one true answer.

# 8. Problem 8

## 8.1. Question

In some technological process hree types of indicator are used to signal the failure. Probability [of] applying indicator of the first type is $$\frac{3}{10}$$, of the second type $$0.4$$, of the third time $$0.3$$. Indicators of particular types signal the failure respectively with probability equal to 0.6, 0.6, 0.4. The probability $$p$$ of the event that the failure would be signaled:

1. $$p \in [0.702;1]$$
2. $$p \in [0.594; 0.702)$$
3. $$[0.27;0.594)$$
4. $$[0;0.27)$$

3 is true; the rest is false

It is so, because: $$p = 0.6 \times 0.3 + 0.6 \times 0.4 + 0.4 \times 0.3 = 0.54$$

# 9. Problem 9

## 9.1. Question

Random variable U has uniform distribution on $$[-3,3]$$. Random variable $$Y = \lfloor U \rfloor$$ (where $$\lfloor x \rfloor$$ denotes the gretest integer not exceeding $$x$$) has c.d.f. with exactly:

1. 9 jumping points
2. 6 jumping points
3. 2 jumping points
4. 5 jumping points
5. 3 jumping points

2 is true; the rest is false

# 10. Problem 10

## 10.1. Question

On the probability space $$([-7,7], \mathcal{B}([-7,7]),P_L)$$ where $$P_L$$ is geometric probability, we define a random wariable $$X(\omega)$$.

$$X(\omega) = \begin{cases} 7 + \omega & \text{for \omega \in [-7; -\frac{1}{5}]} \\ 2 & \text{for \omega \in (-\frac{1}{5};\frac{1}{5})} \\ \frac{7}{2}-\omega & \text{for \omega \in [\frac{1}{5}; 7]}\end{cases}$$

The value of c.d.f. of random variable $$X$$ at the point $$2$$ (i.e. $$F_X(2)$$) is equal:

1. $$\frac{47}{140}$$
2. $$\frac{1}{7}$$
3. $$\frac{11}{28}$$
4. $$\frac{103}{140}$$
5. $$\frac{15}{28}$$

This one is tricky, as the result depends on what definition of c.d.f ( $$F_X(x)$$) we choose.
If we choose $$F_X(x) = P(X<x)$$ then we get $$F_X(2) = \frac{15}{28}$$.
If we choose $$F_X(x) = P(X \leq x)$$ then we obtain $$F_X(2) = \frac{79}{140}$$.