Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0 . 4096 and 0 . 2048 respectively. Then the probability of getting exactly 3 successes is equal to :

Question

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0 . 4096 and 0 . 2048 respectively. Then the probability of getting exactly 3 successes is equal to :, 32 625, 80 243, 40 243, 128 625

MARKS App · Accepted Answer

P X = 1 = C 1 5 · p . q 4 = 0 . 4096 P X = 2 = C 2 5 · p 2 · q 3 = 0 . 2048 ⇒ q 2 p = 2 ⇒ q = 4 p and p + q = 1 ⇒ p = 1 5 and q = 4 5 Now P X = 3 = C 3 5 · 1 5 3 · 4 5 2 = 10 × 16 125 × 25 = 32 625

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