The probability that a man fails to hit a tar | vedclass

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(A) Let $p$ be the probability of hitting the target and $q$ be the probability of failing to hit the target. Given $q = \frac{1}{3}$,so $p = 1 - q =

Vedclass · Accepted Answer

$\frac{16}{27}$

Vedclass · Answer

(A) Let $p$ be the probability of hitting the target and $q$ be the probability of failing to hit the target. Given $q = \frac{1}{3}$,so $p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$. The number of trials is $n = 4$. We need to find the probability of hitting the target at least thrice,which is $P(X \geq 3) = P(X = 3) + P(X = 4)$. Using the binomial distribution formula $P(X = k) = {}^nC_k p^k q^{n-k}$: $P(X = 3) = {}^4C_3 \left(\frac{2}{3}ight)^3 \left(\frac{1}{3}ight)^1 = 4 	imes \frac{8}{27} 	imes \frac{1}{3} = \frac{32}{81}$. $P(X = 4) = {}^4C_4 \left(\frac{2}{3}ight)^4 \left(\frac{1}{3}ight)^0 = 1 	imes \frac{16}{81} 	imes 1 = \frac{16}{81}$. Therefore,$P(X \geq 3) = \frac{32}{81} + \frac{16}{81} = \frac{48}{81} = \frac{16}{27}$.

The probability that a man fails to hit a target is $\frac{1}{3}$. If he fires $4$ times,then the probability that he hits the target at least thrice is

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