Practicing Success
A, B and C are three mutually exclusive and exhaustive events. P(A) = 2P(B) = 6P(C). Find P(B). |
0.3 0.5 0.75 0.7 |
0.3 |
P(A) = 2P(B) = 6P(C) When event are mutually exclusive and exhaustive P(A) + P(B) + P(C) = 1 Let P(A) be k ⇒ P(B) = k/2 ⇒ P(C) = k/6 So according to the concept ⇒ k + k/2 + k/6 = 1 ⇒ 10k/6 = 1 ⇒ k = 3/5 ∴ P(B) is k/2 = 3/(5 × 2) = 0.3 |