51. In a box, there are 250 light bulbs, of which 100 are 100W, 50 are 60W, 50 are 25W, and 50 are 15W. Calculate the probability that the power of any randomly selected light bulb will not exceed 60W.
Solution. Let $A$ be the event that the power of the bulb is $60 \mathrm{Wt}$, $B$-- $25 \mathrm{Wt}$, $C$-- $15 \mathrm{Wt}$, $D$-- $100$ Wt. Events $A, B, C$ and $D$ form a complete system, as they are all mutually exclusive and one of them will definitely occur in the given trial (selection of a bulb). The probability of one of them occurring is a certain event, i.e., $P(A)+P(B)+P(C)+P(D)=1$.

Events "the power of the bulb is no more than 60 Wt" and "the power of the bulb is more than $60 \mathrm{Wt}$" are opposite. According to the property of opposite events, we have $P(A)+P(B)+P(C)=1-P(D)$. Considering that $P(A)+P(B)+P(C)=P(A+B+C)$, we get $P(A+B+C)=1-$ $-P(D)$, hence $P(A+B+C)=1-\frac{100}{250}$, i.e., $P(A+B+C)=\frac{3}{5}$.