The probability that a man can hit a target i | vedclass

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(D) This is a binomial distribution problem where $n = 5$,$p = \frac{3}{4}$,and $q = 1 - p = \frac{1}{4}$.Let $X$ be the number of times he hits the t

Vedclass · Accepted Answer

$\frac{459}{512}$

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(D) This is a binomial distribution problem where $n = 5$,$p = \frac{3}{4}$,and $q = 1 - p = \frac{1}{4}$.Let $X$ be the number of times he hits the target. We need to find $P(X \ge 3) = P(X=3) + P(X=4) + P(X=5)$.Using the formula $P(X=k) = {}^nC_k p^k q^{n-k}$:$P(X=3) = {}^5C_3 (\frac{3}{4})^3 (\frac{1}{4})^2 = 10 	imes \frac{27}{64} 	imes \frac{1}{16} = \frac{270}{1024}$.$P(X=4) = {}^5C_4 (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 	imes \frac{81}{256} 	imes \frac{1}{4} = \frac{405}{1024}$.$P(X=5) = {}^5C_5 (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 	imes \frac{243}{1024} 	imes 1 = \frac{243}{1024}$.Summing these probabilities: $P(X \ge 3) = \frac{270 + 405 + 243}{1024} = \frac{918}{1024} = \frac{459}{512}$.

The probability that a man can hit a target is $\frac{3}{4}$. He tries $5$ times. The probability that he will hit the target at least three times is

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