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(D) Let $p$ be the probability of a case being solved,so $p = 25\% = \frac{1}{4}$.Let $q$ be the probability of a case not being solved,so $q = 1 - p

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$\frac{19}{4096}$

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(D) Let $p$ be the probability of a case being solved,so $p = 25\% = \frac{1}{4}$.Let $q$ be the probability of a case not being solved,so $q = 1 - p = \frac{3}{4}$.Given $n = 6$ trials,we use the binomial distribution formula $P(X=r) = {^nC_r} p^r q^{n-r}$.We need the probability that at least $5$ cases are solved,which is $P(X \ge 5) = P(X=5) + P(X=6)$.$P(X=5) = {^6C_5} \left(\frac{1}{4}ight)^5 \left(\frac{3}{4}ight)^1 = 6 	imes \frac{1}{1024} 	imes \frac{3}{4} = \frac{18}{4096}$.$P(X=6) = {^6C_6} \left(\frac{1}{4}ight)^6 \left(\frac{3}{4}ight)^0 = 1 	imes \frac{1}{4096} 	imes 1 = \frac{1}{4096}$.Therefore,$P(X \ge 5) = \frac{18}{4096} + \frac{1}{4096} = \frac{19}{4096}$.

It is observed that $25\%$ of the cases related to child labour reported to the police station are solved. If $6$ new cases are reported,then the probability that at least $5$ of them will be solved is . . . . . .

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