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Question
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P(A ∩ B) = `1/6`, P(B ∩ C) = `1/4`, P(A ∩ C) = `1/8`, P(A ∪ B ∪ C) = `9/10` and P(A ∩ B ∩ C) = `1/15`, then find P(A), P(B) and P(C)
Solution
By the given condition,
P(B) = 2P(A), P(C) = 3P(A)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)
`9/10 = "P"("A") + 2"P"("A") + 3"P"("A") - 1/6 - 1/4 - 1/8 + 1/15`
`9/10 = 6"P"("A") - 1/6 - 1/4 - 1/8 + 1/15`
6P(A) = `9/10 + 1/6 + 1/4 + 1/8 - 1/15`
= `(108 + 20 + 30 + 15 - 8)/120`
= `(173 - 8)/120`
= `165/120`
= `33/24`
= `11/8`
6P(A) = `11/8`
⇒ P(A) = `11/(6 xx 8)`
P(A) = `11/48`
P(B) = 2 × P(A)
P(B) = `2 xx 11/48 = 11/24`
P(C) = 3P(A)
= `3 xx 11/48 = 11/6`
P(A) = `11/48`, P(B) = `11/24`, P(C) = `11/16`
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