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The mean and variance of a random variable X having a binomial distribution are `4 and 2` respectively. The `P(X=1)` is
A. `1//4`
B. `1//32`
C. `1//16`
D. `1//8`

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Correct Answer - B
Let n and p be the parameters of the binomial distribution.
We have,
Mean `=4` and Variance `=2`
`rArr np=4 and npq=2`
`rArr p=1//2 =q and n=8`
`therefore` Required Probability `=P(X-1)=.^8C_(1)((1)/(2))^1((1)/(2))^7=(1)/(32)`

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