# 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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