Question
Download Solution PDFIf the probabilities of four mutually exclusive and exhaustive events P, Q, R and S satisfy the relation
3P(P) = P(Q) = 2P(R) = 4P(S), then P(Q) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFIf the events P, Q, R, and S are mutually exclusive and exhaustive, then their probabilities must sum to 1. That is: P(P) + P(Q) + P(R) + P(S) = 1
We are given that: 3P(P) = P(Q) = 2P(R) = 4P(S)
We can rewrite this as:
P(P) = \(\frac{P(Q)}{3}\), P(R) = \(\frac{P(Q)}{2}\)and P(S) = \(\frac{P(Q)}{4} \)
Substituting these values into the equation P(P) + P(Q) + P(R) + P(S) = 1, we get:
\(\frac{P(Q)}{3} + P(Q) + \frac{P(Q)}{2} + \frac{P(Q)}{4} = 1\)
Or, \((\frac{1}{3} + 1 + \frac{1}{2} + \frac{1}{4}) \times P(Q) = 1\)
Or, \(\frac{25}{12} P(Q) = 1\)
Or, P(Q) = \(\frac{12}{25}\)
Last updated on Jun 25, 2025
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