108. Two out of three independently operating elements of a computing device have failed. Find the probability that the first and second elements have failed, if the probabilities of failure of the first, second, and third elements are 0.2; 0.4 and 0.3, respectively.
Solution. Let $A$ be the event that two elements have failed. The following assumptions (hypotheses) can be made:

$B_{1}$ - the first and second elements have failed, while the third element is intact, and (since the elements operate independently, the multiplication theorem applies)

$$
P\left(B_{1}\right)=p_{1} \cdot p_{2} \cdot q_{3}=0.2 \cdot 0.4 \cdot 0.7=0.056
$$

$B_{2}$ - the first and third elements have failed, while the second element is intact, and

$$
P\left(B_{2}\right)=p_{1} \cdot p_{3} \cdot q_{2}=0.2 \cdot 0.3 \cdot 0.6=0.036
$$

$B_{3}$ - the second and third elements have failed, while the first is intact, and

$$
P\left(B_{3}\right)=p_{2} \cdot p_{3} \cdot q_{1}=0.4 \cdot 0.3 \cdot 0.8=0.096
$$

$B_{4}$ - only one element has failed; $B_{5}$ - all three elements have failed; $B_{8}$ - none of the elements have failed.

The probabilities of the last three hypotheses have not been calculated, as under these hypotheses the event $A$ (two elements have failed) is impossible, and thus the conditional probabilities $P_{B_{1}}(A), P_{B_{5}}(A)$, and $P_{B_{0}}(A)$ are zero, which means the products $P\left(B_{4}\right) \cdot P_{B_{4}}(A), P\left(B_{5}\right) \times$ $\times P_{B_{5}}(A)$ and $P\left(B_{6}\right) \cdot P_{B_{9}}(A)$ [see the relation (*) below] are zero for any values of the probabilities of hypotheses $B_{4}, B_{5}$, and $B_{6}$.

Since under hypotheses $B_{1}, B_{2}, B_{3}$ the event $A$ is certain, the corresponding conditional probabilities are equal to one:

$$
P_{B_{1}}(A)=P_{B_{2}}(A)=P_{B_{3}}(A)=1
$$

By the formula of total probability, the probability that two elements have failed is

$$
\begin{gathered}
P(A)=P\left(B_{1}\right) \cdot P_{B_{1}}(A)+P\left(B_{2}\right) \cdot P_{B_{2}}(A)+P\left(B_{3}\right) \cdot P_{B_{2}}(A)+ \\
+P\left(B_{4}\right) \cdot P_{B_{1}}(A)+P\left(B_{5}\right) \cdot P_{B_{5}}(A)+P\left(B_{6}\right) \cdot P_{B_{0}}(A)= \\
=0.056 \cdot 1+0.036 \cdot 1+0.096 \cdot 1=0.188 .
\end{gathered}
$$

By Bayes' formula, the required probability that the first and second elements have failed is

$$
P_{A}\left(B_{1}\right)=\frac{P\left(B_{1}\right) \cdot P_{B_{1}}(A)}{P(A)}=\frac{0.056}{0.188}=0.3
$$

109*. Two out of four independently operating lamps of an instrument have failed. Find the probability that the first and second lamps have failed, if the probabilities of failure of the first, second, third, and fourth lamps are respectively: $p_{1}=0.1, p_{2}=0.2, p_{3}=0.3$ and $p_{4}=0.4$.

## Chapter Three

## REPETITION OF EXPERIMENTS

## § 1. Bernoulli's Formula

If trials are conducted in which the probability of the occurrence of an event $A$ in each trial does not depend on the outcomes of other trials, then such trials are called independent with respect to event $A$. In $\S 1-4$ of this chapter, independent trials are considered, in each of which the probability of the occurrence of the event is the same.

Bernoulli's Formula. The probability that in $n$ independent trials, in each of which the probability of the occurrence of an event is $\rho(0<p<1)$, the event will occur exactly $k$ times (regardless of the sequence), is

$$
P_{n}(k)=C_{n}^{k} p^{k} q^{n-k}
$$

or

$$
P_{n}(k)=\frac{n!}{k!(n-k)!} p^{k} q^{n-k}
$$

where $q=1-p$.

The probability that in $n$ trials the event will occur: a) fewer than $k$ times; b) more than $k$ times; c) at least $k$ times; d) at most $k$ times, is found respectively by the formulas:

$$
\begin{gathered}
P_{n}(0)+P_{n}(1)+\ldots+P_{n}(k-1) ; \\
P_{n}(k+1)+P_{n}(k+2)+\ldots+P_{n}(n) ; \\
P_{n}(k)+P_{n}(k+1)+\ldots+P_{n}(n) ; \\
P_{n}(0)+P_{n}(1)+\ldots+P_{n}(k)
\end{gathered}
$$