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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 $q = \f

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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 $q = \frac{npq}{np} = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n(\frac{1}{2}) = 2$,which implies $n = 4$. We need to find the probability $P(X \geq 1)$. Using the complement rule,$P(X \geq 1) = 1 - P(X = 0)$. The probability mass function is $P(X = k) = {}^nC_k p^k q^{n-k}$. For $k = 0$,$P(X = 0) = {}^4C_0 (\frac{1}{2})^0 (\frac{1}{2})^4 = 1 	imes 1 	imes \frac{1}{16} = \frac{1}{16}$. Therefore,$P(X \geq 1) = 1 - \frac{1}{16} = \frac{15}{16}$.

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 $1$ is

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