Problem 8. In 28 examination tickets, each includes two theoretical questions and one problem. The student has prepared 50 theoretical questions and can solve the problems in 22 tickets. What is the probability that, by randomly picking one ticket, the student will be able to answer all the questions in the ticket?
Solution. Consider the events:

$A_{1}$ - the student knows the first question;

$A^{2}$ - the student knows the second question;

$A_{3}$ - the student will solve the problem;

$A=A_{1} A_{2}$ - the student knows both questions;

Then $A A_{3}$ - the student will answer all the questions on the ticket. Events $A_{1}$ and $A_{2}$ are dependent, $A$ and $A_{3}$ are independent.

We have $p\left(A_{1}\right)=\frac{50}{56} ; p\left(A_{2} / A_{1}\right)=\frac{49}{55} ; p\left(A_{3}\right)=\frac{22}{28}=\frac{11}{14}$.

By formulas (10) and (9) we get:

$$
\begin{aligned}
p\left(A A_{3}\right)=p(A) p\left(A_{3}\right)=p\left(A_{1}\right) p\left(A_{2} / A_{1}\right) p\left(A_{3}\right) & = \\
& =\frac{50}{56} \cdot \frac{49}{55} \cdot \frac{22}{28}=\frac{5}{8}=0.625
\end{aligned}
$$

Answer. $0.625$.

## Exercises

1. A salvo is fired from two guns at a target. The probability of hitting with the first gun is 0.85; with the second 0.91. Find the probability of hitting the target.
2. In a lottery, 24 monetary and 10 material prizes are distributed among 1000 tickets. What is the probability of winning with two tickets: a) at least one ticket; b) one monetary prize and one material prize?
3. A company produces $95\%$ of standard products, of which $86\%$ are of the first grade. Find the probability that a randomly selected product, produced by this company, will be of the first grade.
4. In an urn, there is 1 black and 9 white balls. Three balls are drawn at once. What is the probability that all the drawn balls are white?
5. Three shots are fired at a target. The probability of hitting with each shot is 0.5. Find the probability that only one hit will occur as a result of these shots.
6. In the first urn, there is 1 white, 2 red, and 3 blue balls. In the second urn, there are 2 white, 6 red, and 4 blue balls. One ball is drawn from each urn. What is the probability that none of them are blue?
7. Among 20 electric bulbs, 3 are non-standard. Find the probability that two bulbs taken sequentially will be non-standard.
8. A student intends to take a needed book from one of two libraries. The probability that the book is in the first library is 0.75; in the second 0.85. Find the probability of the events: a) the book will be found only in one library; b) at least in one library.
9. Two shooters fire at a target. The probability of hitting with one shot for the first shooter is 0.75; for the second 0.85. Find the probability that in one salvo at the target: a) only one of the shooters will hit; b) only the first shooter will hit; c) both shooters will hit; d) neither will hit.
10. In an urn, there are 9 white and 3 red balls. Three balls are drawn sequentially without replacement. Find the probability that all the balls are white.
11. Four telephone sets reliably operate during the day with probabilities of 0.3, 0.4, 0.4, and 0.7, respectively. Find the probability that during the day: a) all 4 sets will operate reliably; b) only two sets will operate reliably; c) only one set will operate reliably; d) none will operate reliably.

## Answers

1. 0.9865. 2. a) 0.067; b) 0.00024. 3. 0.817. 4. 0.7. 5. 0.375. 6. $\frac{1}{3}$. 7. $\frac{3}{190}$. 8. a) 0.325; b) 0.9625. 9. a) 0.325; b) 0.1125; c) 0.6375; d) 0.0375. 10. $\frac{21}{35}=0.3818$. 11. a) 0.0336; b) 0.3876; c) 0.3096; d) 0.0756.

## § 5. Theorem of Addition of Probabilities of Joint Events

Theorem 5. The probability of the occurrence of at least one of two joint events is equal to the sum of the probabilities of these events minus the probability of their joint occurrence:

$$
p(A+B)=p(A)+p(B)-p(A B)
$$

Corollary 1. If event $B$ depends on event $A$, then $p(A B)=p(A) p(B / A)$ and from (13) it follows:

$$
p(A+B)=p(A)+p(B)-p(A) p(B / A)
$$

Corollary 2. If events $A$ and $B$ are independent, then $p(A B)=$ $p(A) p(B)$ and from (13) it follows:

$$
p(A+B)=p(A)+p(B)-p(A) p(B)
$$

Corollary 3. If events $A$ and $B$ are mutually exclusive, then $p(A B)=0$, from which it follows:

$$
p(A+B)=p(A)+p(B)
$$

## Problems with Solutions