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(D) Given the probability of success $p = 3/4$,the probability of failure $q = 1 - 3/4 = 1/4$,and the number of trials $n = 5$. We need to find the pr

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$459/512$

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(D) Given the probability of success $p = 3/4$,the probability of failure $q = 1 - 3/4 = 1/4$,and the number of trials $n = 5$. We need to find the probability of at least $3$ successes,which is $P(X \ge 3) = P(X=3) + P(X=4) + P(X=5)$. Using the binomial distribution formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$: $P(X=3) = \binom{5}{3} (3/4)^3 (1/4)^2 = 10 	imes (27/64) 	imes (1/16) = 270/1024$. $P(X=4) = \binom{5}{4} (3/4)^4 (1/4)^1 = 5 	imes (81/256) 	imes (1/4) = 405/1024$. $P(X=5) = \binom{5}{5} (3/4)^5 (1/4)^0 = 1 	imes (243/1024) 	imes 1 = 243/1024$. Summing these probabilities: $P(X \ge 3) = (270 + 405 + 243) / 1024 = 918 / 1024$. Simplifying the fraction: $918 / 1024 = 459 / 512$.

The probability of a man hitting a target is $3/4$. He tries $5$ times. What is the probability that he hits the target at least $3$ times?

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