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

Solution :

Given that, mean = 4 \(\implies\) np = 4

And Variance = 2 \(\implies\) npq = 2 \(\implies\) 4q = 2

\(\implies\)  q = \(1\over 2\)

\(\therefore\)   p = 1 – q = 1 – \(1\over 2\) = \(1\over 2\)

Also, n = 8

Probability of 2 successes = P(X = 2) = \(^8C_2\)\(p^2\)\(q^6\)

= \(8!\over {2!\times 6!}\) \(\times\) \(({1\over 2})^2\) \(\times\) \(({1\over 2})^2\)

= \(28\over 256\)


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