1523. The master services 5 machines. $20 \%$ of the working time he spends at the first machine, 10 - at the second, 15 - at the third, $25-$ at the fourth and, finally, 30 - at the fifth. Find the probability that at a randomly chosen moment in time he is: 1) at the first or third machine; 2) at the second or fifth; 3) at the first or fourth; 4) at the first or second, or third; 5) at the fourth or fifth.
Solution. Let's introduce the notation. Let $A, B, C, D, E$ be the events that occur if the master is at the first, second, third, fourth, and finally, the fifth machine at a randomly chosen moment in time, respectively. From the problem statement, it follows that $A, B, C, D, E$ are pairwise mutually exclusive events and $P(A)=0.20; P(B)=0.10; P(C)=0.15; P(D)=0.25; P(E)=0.30$.

1) The event that occurs if the master is at the first or third machine is $A+C$. We need to find $P(A+C)$. By the theorem of the addition of probabilities,

$P(A+C)=P(A)+P(C)=0.20+0.15=0.35$.

2) $P(B+E)=P(B)+P(E)=0.10+0.30=0.40$.

3) $P(A+D)=P(A)+P(D)=0.20+0.25=0.45$.

4) $P(A+B+C)=P(A)+P(B)+P(C)=0.20+0.10+$ $+0.15=0.45$.

5) $P(D+E)=P(D)+P(E)=0.25+0.30=0.55$.