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(D) For a binomial distribution,the mean is given by $np = 2$ and the variance is given by $npq = 1$.Dividing the variance by the mean,we get $\frac{n

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$\frac{15}{16}$

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(D) For a binomial distribution,the mean is given by $np = 2$ and the variance is given by $npq = 1$.Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{1}{2}$,which implies $q = \frac{1}{2}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n 	imes \frac{1}{2} = 2$,so $n = 4$.The probability that $X$ takes a value greater than or equal to one is $P(X \ge 1) = 1 - P(X = 0)$.Using the binomial probability formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$,we have $P(X = 0) = \binom{4}{0} (\frac{1}{2})^0 (\frac{1}{2})^4 = 1 	imes 1 	imes \frac{1}{16} = \frac{1}{16}$.Therefore,$P(X \ge 1) = 1 - \frac{1}{16} = \frac{15}{16}$.

If the mean and the variance of a binomial variate $X$ are $2$ and $1$ respectively,then the probability that $X$ takes a value greater than or equal to one is

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