The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is - Mathematics MCQ AIEEE 2004 2026 | ExamTest.live | ExamTest.live

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The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

A $\frac{37}{256}$
B $\frac{219}{256}$
C $\frac{128}{256}$
D $\frac{28}{256}$

Solution & Step-by-step Explanation

For a binomial distribution: Mean n p = 4 . Variance n pq = 2 . Dividing variance by mean: q = 2/4 = 1/2 . Then p = 1 - q = 1/2 . Since n p = 4 and p = 1/2, then n (1/2) = 4 \Rightarrow n = 8 . Probability of 2 successes (x = 2): P (x = 2) =^{8} C_{2} \cdot (1/2)^{2} \cdot (1/2)^{8 - 2} =^{8} C_{2} \cdot (1/2)^{8} .^{8} C_{2} = \frac{8 \times 7}{2 \times 1} = 28 . P (x = 2) = 28 \cdot \frac{1}{256} = \frac{28}{256} .

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The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

A

\frac{37}{256}

B

\frac{219}{256}

C

\frac{128}{256}

D

\frac{28}{256}

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