The probability that a bomb will miss the tar | vedclass

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Question

(B) The probability of missing the target is $q = 0.2 = \frac{1}{5}$.So,the probability of hitting the target is $p = 1 - q = 1 - 0.2 = 0.8 = \frac{4}

Vedclass · Accepted Answer

$\frac{144}{5^{9}}$

Vedclass · Answer

(B) The probability of missing the target is $q = 0.2 = \frac{1}{5}$.So,the probability of hitting the target is $p = 1 - q = 1 - 0.2 = 0.8 = \frac{4}{5}$.Given $n = 10$ bombs are dropped,and we want exactly $r = 2$ hits.Using the binomial distribution formula $P(X = r) = {}^{n}C_{r} p^{r} q^{n-r}$:$P(X = 2) = {}^{10}C_{2} 	imes (0.8)^{2} 	imes (0.2)^{8}$$P(X = 2) = \frac{10 	imes 9}{2 	imes 1} 	imes \left(\frac{4}{5}ight)^{2} 	imes \left(\frac{1}{5}ight)^{8}$$P(X = 2) = 45 	imes \frac{16}{25} 	imes \frac{1}{5^{8}}$$P(X = 2) = 45 	imes \frac{16}{5^{2} 	imes 5^{8}} = 45 	imes \frac{16}{5^{10}}$$P(X = 2) = (9 	imes 5) 	imes \frac{16}{5^{10}} = \frac{9 	imes 16}{5^{9}} = \frac{144}{5^{9}}$Thus,the correct option is $B$.

The probability that a bomb will miss the target is $0.2$. Then the probability that out of $10$ bombs dropped,exactly $2$ will hit the target is:

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