A man makes attempts to hit the target. The p | vedclass

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(A) This is a binomial distribution problem where $n = 5$ (number of attempts),$p = \frac{3}{5}$ (probability of success),and $q = 1 - p = \frac{2}{5}

Vedclass · Accepted Answer

$\frac{144}{625}$

Vedclass · Answer

(A) This is a binomial distribution problem where $n = 5$ (number of attempts),$p = \frac{3}{5}$ (probability of success),and $q = 1 - p = \frac{2}{5}$ (probability of failure).The probability of exactly $k$ successes in $n$ trials is given by the formula $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$.For $k = 2$:$P(X = 2) = {}^5C_2 \cdot \left(\frac{3}{5}\right)^2 \cdot \left(\frac{2}{5}\right)^{5-2}$$P(X = 2) = 10 \cdot \left(\frac{9}{25}\right) \cdot \left(\frac{8}{125}\right)$$P(X = 2) = 10 \cdot \frac{72}{3125} = \frac{720}{3125} = \frac{144}{625}$.

$A$ man makes attempts to hit the target. The probability of hitting the target is $\frac{3}{5}$. Then the probability that he hits the target exactly $2$ times in $5$ attempts is:

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