Class 12
MATHS
Binomial Distribution

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

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Text Solution

(28)/(256)(219)/(256)(128)/(256)(37)/(256)

A
Solution :
Let n and p be the parameters for the distribution . Then <br> np=4 and npq=2 <br> implies p=q=1/2 and n=8 <br> let X denotes the number of successes. Then , <br> P(X-r)=""^(8)C_(r)((1)/(2))^(r)((1)/(2))^(8-r)=""^(8)C_(r)((1)/(2))^(8) <br> therefore  Required probability =P(X=2)=""^(8)C_(2)((1)/(2))^(8)=(28)/(256)

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