45. How many numbers need to be selected from a table of random numbers to be confident with a probability of at least 0.9 that among them there is at least one even number?
Solution. Let $n$ be the required number of random numbers. Consider the events:

$A_{k}$ - one randomly selected number is even $(k=1,2, \ldots, n)$;

$\overline{A_{k}}$ - one randomly selected number is odd;

$B$ - among $n$ random numbers there will be at least one even;

$\bar{B}$ - among $n$ random numbers there will be no even numbers.

The probabilities of all events $\overline{A_{k}}$ are equal. Let $P\left(A_{k}\right)=p$, $P\left(\bar{A}_{k}\right)=1-p=q$. Events $A_{k}$ and $\overline{A_{k}}$ are equally probable, so $p=q=0.5$. The probability of event $B$ is

$$
P(B)=1-P(\bar{B})
$$

Event $\bar{B}$ consists in all $n$ random numbers being odd. This means that $\bar{B}$ is the product of $\boldsymbol{n}$ independent events $\overline{A_{k}}: \bar{B}=\overline{A_{1}} \cdot \overline{A_{2}} \cdot \ldots \cdot \overline{A_{n}}$.

Applying the multiplication theorem of probabilities of independent events and equality $(*)$, we find $P(B)=1-q^{n}=1-0.5^{n}$.

By the condition $P(B) \geq 0.9$, we get: $1-0.5^{n} \geq 0.9$.

Solving this inequality, we sequentially find:

$$
\begin{gathered}
-0.5^{n} \geq-0.1 \\
0.5^{n} \leq 0.1 \\
n \cdot \lg 0.5 \leq \lg 0.1
\end{gathered}
$$

Since $\lg 0.5(\lg 0.1 / \lg 0.5) \approx 3.32$. Since $n$ is an integer, then $\boldsymbol{n} \geq 4$.

## 2.2. The Formula of Total Probability

The probability of an event $A$, which can occur given the occurrence of one of $n$ mutually exclusive events (hypotheses) $B_{i}(i=1,2, \ldots, n)$, forming a complete group, is found using the formula of total probability:

$$
P(A)=\sum_{i=1}^{n} P\left(B_{i}\right) P_{B_{i}}(A)
$$

In this formula, $P(A)$ is the probability of event $A$; $P\left(B_{i}\right)$ is the probability of event $B_{i}$, $P_{B_{i}}(A)$ is the conditional probability of event $A$, calculated given that event $B_{i}$ has occurred; $\sum_{i=1}^{n} P\left(B_{i}\right) P_{B_{i}}(A)$ is the sum of the products of the probabilities $P\left(B_{i}\right)$ of each of the events $B_{i}$ and the corresponding conditional probability $P_{B_{l}}(A)$.

The sum of the probabilities of the hypotheses $\sum_{i=1}^{n} P\left(B_{i}\right)=1$.