If the three independent events E1, E2 and E3, the probability that on

1.	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» (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.

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»

(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.