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

Question

If the three independent events E₁, E₂ and E₃, the probability that only E₁ occurs is \(\alpha\), E₂ occurs is \(\beta\), only E₃ occurs is \(\gamma\). Let the probability p that none of the event E₁, E₂ or E₃ occurs satisfy the equations (\(\alpha\) – 2\(\beta\)) p = \(\alpha\)\(\beta\) and (\(\beta\) – 3\(\gamma\)) p = 2\(\beta\)\(\gamma\). All the given probabilities are assumed to lie in the interval (0, 1).

Then, \(\frac{\text{Probability of occurrence of}\,E_1}{\text{Probability of occurrence of}\,E_3}\) is equal to

(a) 3 

(b) 5 

(c) 6 

(d) 8

Answer 1

(c) 6

Let \(x\), 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))

⇒ \(\frac{x}{z}\) = 6 ⇒ \(\frac{P(E_1)}{P(E_3)}\) = 6.