JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Probability

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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 :

(A) $\frac{40}{243}$
(B) $\frac{80}{243}$
(C) $\frac{32}{625}$ ✓ Correct
(D) $\frac{128}{625}$

Answer: (C)

Solution

With

n = 5

and success probability

p

, the two given probabilities are

P (X = 1) = 5 p q^{4} = 0.4096

and

P (X = 2) = 10 p^{2} q^{3} = 0.2048

, where

q = 1 - p

. Dividing the second by the first,

\frac{10 p ^{2} q ^{3}}{5 p q ^{4}} = \frac{2 p}{q} = \frac{0.2048}{0.4096} = \frac{1}{2}

q = 4 p

, and

p + q = 1

gives

p = \frac{1}{5}

q = \frac{4}{5}

. Then

P (X = 3) = (3 5) p^{3} q^{2} = 10 \cdot \frac{1}{125} \cdot \frac{16}{25} = \frac{160}{3125} = \frac{32}{625}

so the probability of exactly three successes is

\frac{32}{625}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Probability

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