Problem 3. An inspector checks items for compliance with the standard. It is known that the probability of an item complying with the standard is 0.95. Find the probability that: a) both of the two inspected items will be standard, if the events of items complying with the standard are independent; b) only one of the two inspected items is standard.
Solution. a) Let $A_{1}$ and $A_{2}$ be events indicating that the first and second items are standard, respectively. Then $p\left(A_{1}\right)=p\left(A_{2}\right)$.

Therefore, $p\left(A_{1} A_{2}\right)=p\left(A_{1}\right) p\left(A_{2}\right)=0.95 \cdot 0.95=0.9025$ b) If $B$ is the event indicating that only one of the two items is standard, then $B=A_{1} \overline{A_{2}}+\overline{A_{1}} A_{2}$. In this case, $p\left(\overline{A_{1}}\right)=p\left(\overline{A_{2}}\right)=0.05-$ is the probability of non-standardness of the item. Since $A_{1} \overline{A_{2}}$ and $\overline{A_{1}} A_{2}$ are mutually exclusive, then $p(B)=2 \cdot 0.95 \cdot 0.05=0.095$.

Answer. a) 0.9025; b) 0.095.

## Exercises

1. The probability of timely delivery of newspapers to each of two post offices is 0.9. Find the probability that at least one post office will receive the newspapers on time.
2. What is the probability that 6 points will appear on at least one die when two dice are rolled?
3. Find the probability that a randomly selected two-digit number will be divisible by either 2, or 5, or both simultaneously.
4. Find the probability that two doubles will be drawn in a row from a complete set of dominoes.

## Answers

1. $0.99 \cdot$ 2. $\frac{11}{36}=0.3056$. 3. $0.6 \cdot$ 4. $\frac{1}{18}=0.0556$.

## § 6. The Formula of Total Probability. Bayes' Formula

Theorem 6 (on total probability). Let event $A$ occur only together with any of the events $H_{1}, H_{2}, \ldots, H_{n}$, forming a complete group of events. Then event $A$ can be represented as the sum $A=A H_{1}+A H_{2}+\ldots+A H_{n}$, and its probability is determined by the formula of total probability:

$$
p(A)=p\left(H_{1}\right) p\left(A / H_{1}\right)+p\left(H_{2}\right) p\left(A / H_{2}\right)+\ldots+p\left(H_{n}\right) p\left(A / H_{n}\right)
$$

In this case, the events $H_{1}, H_{2}, \ldots, H_{n}$ are called hypotheses, and their probabilities are called the probabilities of the hypotheses.

Bayes' Formula. Suppose that in the conditions of the theorem on total probability, one trial is performed, as a result of which event $A$ occurs. In such a case, the prior (or a priori) probabilities of the hypotheses $p\left(H_{1}\right), p\left(H_{2}\right), \ldots, p\left(H_{n}\right)$ should be replaced by new, posterior (or a posteriori) probabilities, which are calculated using Bayes' formula:

$$
p\left(H_{k} / A\right)=\frac{p\left(H_{k}\right) p\left(A / H_{k}\right)}{p(A)} \quad(k=1,2, \ldots, n)
$$

where the probability $p(A)$ is determined by the formula of Theorem 6.

Bayes' formula follows from the assumption that event $A$ occurred simultaneously with one of the hypotheses, for example, $H_{k}$. In such a case,

$$
p\left(A H_{k}\right)=p(A) p\left(H_{k} / A\right)=p\left(H_{k}\right) p\left(A / H_{k}\right)
$$

## Problems with Solutions