The mean and variance of a random variable having a binomial distribution are 4 and 2 respectively, then P(X = 1) is: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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The mean and variance of a random variable having a binomial distribution are 4 and 2 respectively, then P (X = 1) is:

Solution & Step-by-step Explanation

For a binomial distribution: Mean n p = 4 Variance n pq = 2 Dividing the two: q = \frac{2}{4} = \frac{1}{2} ⟹ p = 1 - q = \frac{1}{2} . From n p = 4 ⟹ n (1/2) = 4 ⟹ n = 8 . Now, P (X = 1) =^{8} C_{1} p^{1} q^{8 - 1} $ P (X = 1) = 8 (\frac{1}{2}) (\frac{1}{2})^{7} = \frac{8}{2 ^{8}} = \frac{8}{256} = \frac{1}{32} $

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The mean and variance of a random variable having a binomial distribution are 4 and 2 respectively, then P (X = 1) is:

\frac{1}{32}

\frac{1}{16}

\frac{1}{8}

\frac{1}{4}

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