Question

If the three independent events E1, E2 and E3, the probability that only E1 occurs is α, E2 occurs -Maths 9th

Answer 1

(c) 6Let \(x\), y, z be the probabilities of happening of events E1, E2 and E3 respectively. Then, \(\alpha\) = P (occurrence of E1 only) = x (1 – y) (1 – z) \(\beta\) = P (occurrence of E2 only) = (1 – \(x\)) y (1 – z) \(\gamma\) = P (occurrence of E3 only) = (1 – \(x\)) (1 – y) z p = P (not occurrence of E1, E2, E3) = (1 – \(x\)) (1 – y) (1 – z). ∴ (\(\alpha\) – 2\(\beta\)) p = \(\alpha\)\(\beta\) ⇒ [\(x\)(1 – y) (1 – z) – 2 (1 – \(x\))y (1 – z)] (1 – \(x\)) (1 – y) (1 – z) = \(x\) (1 – y) (1 – z) (1 – \(x\)) y (1 – z) ⇒ (1 – z) [\(x\) (1 – y) – 2y (1 – \(x\))] = \(x\)y (1 – z) ⇒ \(x\) – \(x\)y – 2y + 2\(x\)y = \(x\)y ⇒ \(x\) = 2y                                ...(i) Also, (\(\beta\) – 3\(\gamma\)) p = 2\(\beta\)\(\gamma\) ⇒ y = 3z             ...(ii)⇒ y = 3z           ...(ii) ∴ \(x\) = 6z (From (i) and (ii))⇒ \(rac{x}{z}\) = 6 ⇒ \(rac{P(E_1)}{P(E_3)}\) = 6.