# 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:

- \(P(Z>=0) > P(W>=0)\)
- \(P(Z>0) = P(W>0)\)
- \(P(Z>0) = P(W<0)\)
- \(P(Z<0) = P(W<0)\)

## 1.2. Answer

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:

- \(\frac{1}{16}\)
- \(1/3\)
- \(3/4\)
- \(3/16\)

## 2.2. Answer

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 \(P(Y= -1) = 1 - P(Y=1) = 1/3\).

- \(V(X-Y) \in (0.5; 2.1222]\)
- \(V(X-Y) > V(X+Y)\)
- \(V(X-Y) >= \frac{29}{9}\)
- \(V(X-Y) <= \frac{20}{9}\)
- \(V(X-Y) >= 3.7222\)

## 3.2. Answer

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:

- \(E\) and \(K\) exclude themselves.
- \(E^C\) and \(K^C\) exclude themselves.
- \(E\) and \(K^C\) can be dependent.
- \(E \cap K\) is an impossible event.
- \(E^C\) and \(K^C\) are independent.

## 4.2. Answer

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:

- \(P(E \cap K) < P(E|K)P(K)\)
- \(P(E \cup K) > P(E)+P(K)\)
- \(P(K \cap E) > min(P(E), P(K))\)
- \(P(E \cup K) < max(P(E),P(K))\)
- \(P(E^C)+P(K^C)+P(E \cup K) - P(E^C \cup K^C) = 1\)

## 5.2. Answer

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 \((0,3)\).

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

## 6.2. Answer

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:

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

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

## 7.2. Answer

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:

- \(p \in [0.702;1]\)
- \(p \in [0.594; 0.702)\)
- \([0.27;0.594)\)
- \([0;0.27)\)

## 8.2. Answer

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:

- 9 jumping points
- 6 jumping points
- 2 jumping points
- 5 jumping points
- 3 jumping points

## 9.2. Answer

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:

- \(\frac{47}{140}\)
- \(\frac{1}{7}\)
- \(\frac{11}{28}\)
- \(\frac{103}{140}\)
- \(\frac{15}{28}\)

## 10.2. Answer

5 is true; the rest is false

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 \(F_X(2) = \frac{79}{140}\).