Let X sim B(n, p). If E(X) = 2 and Var(X) = 1 | vedclass

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(C) For a binomial distribution $X \sim B(n, p)$, we have $E(X) = np = 2$ and $Var(X) = npq = 1$.Dividing the variance by the mean: $q = \frac{npq}{np

Vedclass · Accepted Answer

$0.3125$

Vedclass · Answer

(C) For a binomial distribution $X \sim B(n, p)$, we have $E(X) = np = 2$ and $Var(X) = npq = 1$.Dividing the variance by the mean: $q = \frac{npq}{np} = \frac{1}{2} = 0.5$.Since $p + q = 1$, $p = 1 - 0.5 = 0.5$.Substituting $p$ into $np = 2$: $n(0.5) = 2 \implies n = 4$.We need to find $P(X \le 1) = P(X = 0) + P(X = 1)$.$P(X = k) = \binom{n}{k} p^k q^{n-k}$.$P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = 1 	imes 1 	imes 0.0625 = 0.0625$.$P(X = 1) = \binom{4}{1} (0.5)^1 (0.5)^3 = 4 	imes 0.5 	imes 0.125 = 0.25$.$P(X \le 1) = 0.0625 + 0.25 = 0.3125$.

Let $X \sim B(n, p)$. If $E(X) = 2$ and $Var(X) = 1$, then the probability of getting at most one success is .....

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