46. Electric lamps are manufactured at three plants. The first plant produces $35 \%$ of the total number of lamps, the second - $50 \%$, and the third $15 \%$. The production of the first plant contains $70 \%$ standard lamps, the second $80 \%$, and the third $90 \%$. The products of all three plants are supplied to the store. What is the probability that

a) a randomly selected lamp is manufactured at the first plant and is standard;

b) a lamp purchased from the store is standard?
S o l u t i o n. Let's consider the events:

$B_{1}$ - a randomly selected lamp is manufactured by the first factory;

$B_{2}$ - a randomly selected lamp is manufactured by the second factory;

$B_{3}$ - a randomly selected lamp is manufactured by the third factory;

$C_{1}$ - a randomly selected lamp is manufactured by the first factory and is standard;

$A$ - a lamp purchased from the store is standard.

It should be noted that the first hypothesis - event $B_{1}$, means that a lamp randomly taken from the total number of lamps manufactured by the first factory can be of any quality, i.e., it can be both standard and non-standard. The other two hypotheses of events $B_{2}$ and $B_{3}$ have a similar meaning.

The probabilities of events $B_{1}, B_{2}$, and $B_{3}$ according to formula (1) are: $P\left(B_{1}\right)=0.35 ; P\left(B_{2}\right)=0.5 ; P\left(B_{3}\right)=0.15$. Events $B_{1}, B_{2}$, and $B_{3}$ are mutually exclusive and form a complete group, the sum of the probabilities of these events equals one: $0.35+0.5+0.15=1$.

From the condition, it follows that $P_{B_{1}}(A)=0.7, P_{B_{2}}(A)=0.8, P_{B_{3}}(A)=0.9$.
a) Event $C_{1}$ consists in the fact that a randomly selected lamp, firstly, is manufactured by the first factory, and, secondly, is standard. This means that event $C_{1}$ is the product of two dependent events $B_{1}$ and $A: C_{1}=B_{1} \cdot A$. The probability of event $C_{1}$ will be found using the theorem of multiplication of probabilities of dependent events. Applying formula (12), we get:

$$
P\left(C_{1}\right)=P\left(B_{1}\right) P_{B_{1}}(A)=0.35 \cdot 0.7=0.245
$$

b) Event $A$ represents the sum of the following three mutually exclusive events:

$C_{1}$ - the lamp is manufactured by the first factory and it is standard;

$C_{2}$ - the lamp is manufactured by the second factory and it is standard;

$C_{3}$ - the lamp is manufactured by the third factory and it is standard.

Each of the events $C_{i}(i=1,2,3)$ represents the product of two dependent events $B_{i}$ and $A$.

Thus, $A=B_{1} \cdot A+B_{2} \cdot A+B_{3} \cdot A$. Applying formula (16), we get

$$
\begin{aligned}
& P(A)=P\left(B_{1}\right) P_{B_{1}}(A)+P\left(B_{2}\right) P_{B_{2}}(A)+P\left(B_{3}\right) P_{B_{3}}(A)= \\
& =0.35 \cdot 0.7+0.5 \cdot 0.8+0.15 \cdot 0.9=0.245+0.4+0.135=0.78
\end{aligned}
$$