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`