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Q.1. Of the three independent events E1, E2 and E3, the probability that only E1 occurs is α, only E2 occurs is β and only E3 occurs is γ. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations (α 2β)p = αβ and (β – 3γ)p = 2bg.
All the given probabilities are assumed to lie in the interval (0, 1). (JEE Adv. 2013)
Ans. Sol. (6) Let P(E1) = x; P(E2) = y, P(E3) = z
P(only E1) = x(1 – y) (1 – z) = α
P(only E2) = (1 – x) y (1 – z) = β
P(only E3) = (1 – x) ( 1 – y) z = g
P(none) = (1 – x) (1 – y) (1 – z) = p.
Now given (α – 2β) p = αβ
⇒ x = 2y an d(β – 3r)p = 2βr ⇒ y = 3z
∴ x = 6z
Hence
Q.2. The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is (JEE Adv. 2015)
Ans. Sol. P(x > 2) > 0.96
⇒ 1 – P(x = 0) – P(x = 1) > 0.96
⇒ P(x = 0) + P(x = 1) < 0.04
⇒ 
⇒ minimum value of n is 8.
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