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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