A, B, C and D are mutually exclusive and exhaustive events. If 2P(A)=3P(B)=4P(C)=5P(D), then what is 77P(A) equal to?

Explanation

Let 2P(A) = 3P(B) = 4P(C) = 5P(D) = k. Therefore, P(A) = \frac{k}{2}, P(B) = \frac{k}{3}, P(C) = \frac{k}{4}, P(D) = \frac{k}{5}. Since the events are mutually exclusive and exhaustive, their sum is 1: \frac{k}{2} + \frac{k}{3} + \frac{k}{4} + \frac{k}{5} = 1. The LCM of denominators is 60, so k\left(\frac{30+20+15+12}{60}\right) = 1 \implies k\left(\frac{77}{60}\right) = 1 \implies k = \frac{60}{77}. Thus, P(A) = \frac{60/77}{2} = \frac{30}{77}. We need 77P(A) = 77 \times \frac{30}{77} = 30.

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