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(B) Let $X$ be the number of times the man hits the target. This follows a binomial distribution $B(n, p)$ with $n=8$ and $p=\frac{1}{3}$.Probability

Vedclass · Accepted Answer

$1-5\left(\frac{2}{3}\right)^8$

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(B) Let $X$ be the number of times the man hits the target. This follows a binomial distribution $B(n, p)$ with $n=8$ and $p=\frac{1}{3}$.Probability of success $p = \frac{1}{3}$,probability of failure $q = 1 - \frac{1}{3} = \frac{2}{3}$.We need to find the probability of hitting the target at least twice,which is $P(X \ge 2)$.$P(X \ge 2) = 1 - [P(X=0) + P(X=1)]$.Using the binomial formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:$P(X=0) = \binom{8}{0} (\frac{1}{3})^0 (\frac{2}{3})^8 = (\frac{2}{3})^8$.$P(X=1) = \binom{8}{1} (\frac{1}{3})^1 (\frac{2}{3})^7 = 8 \cdot \frac{1}{3} \cdot (\frac{2}{3})^7$.$P(X \ge 2) = 1 - [(\frac{2}{3})^8 + 8 \cdot \frac{1}{3} \cdot (\frac{2}{3})^7]$.$P(X \ge 2) = 1 - [(\frac{2}{3})^7 \cdot (\frac{2}{3} + \frac{8}{3})] = 1 - [(\frac{2}{3})^7 \cdot \frac{10}{3}] = 1 - 5 \cdot (\frac{2}{3})^8$.

Given that the probability of a man hitting a target with a gun is $\frac{1}{3}$. If he fires $8$ times,then the probability of his hitting the target at least twice is

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