Let A, B, C and D be mutually exclusive and exhaustive events and $\frac{P(A)}{2} = \frac{P(B)}{3} = \frac{P(C)}{5} = \frac{P(D)}{8}$.
To solve the problem, we are given that the events \(A\), \(B\), \(C\), and \(D\) are mutually exclusive and exhaustive. Furthermore, it is given that:
\(\frac{P(A)}{2} = \frac{P(B)}{3} = \frac{P(C)}{5} = \frac{P(D)}{8} = k\)
Here, \(k\) is a constant. From this equation, we can express the probabilities of each event as:
- \(P(A) = 2k\)
- \(P(B) = 3k\)
- \(P(C) = 5k\)
- \(P(D) = 8k\)
Since the events are exhaustive, their probabilities sum to 1:
\(P(A) + P(B) + P(C) + P(D) = 1\)
Substituting the expressions for the probabilities, we have:
\(2k + 3k + 5k + 8k = 1\)
Simplifying this gives:
\(18k = 1\)
Therefore, \(k = \frac{1}{18}\).
Next, we need to find the value of:
\(\frac{[2P(A) + 3P(B)]}{[4P(C) + 5P(D)]}\)
Substituting the values of the probabilities:
\(\text{Numerator} = 2(2k) + 3(3k) = 4k + 9k = 13k\)
\(\text{Denominator} = 4(5k) + 5(8k) = 20k + 40k = 60k\)
Thus, the expression simplifies to:
\(\frac{13k}{60k} = \frac{13}{60}\)
The correct answer is therefore \(\frac{13}{60}\).
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