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(B) The probability of a case being solved is $p = 25 \% = \frac{1}{4}$.Thus,the probability of a case not being solved is $q = 1 - p = \frac{3}{4}$.G

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$\frac{19}{4096}$

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(B) The probability of a case being solved is $p = 25 \% = \frac{1}{4}$.Thus,the probability of a case not being solved is $q = 1 - p = \frac{3}{4}$.Given $n = 6$ trials,we need to find the probability that at least $5$ cases are solved,which is $P(X \ge 5) = P(X = 5) + P(X = 6)$.Using the binomial distribution formula $P(X = x) = {}^nC_x p^x q^{n-x}$:$P(X = 5) = {}^6C_5 \left(\frac{1}{4}ight)^5 \left(\frac{3}{4}ight)^1 = 6 	imes \frac{1}{4^5} 	imes \frac{3}{4} = \frac{18}{4^6}$.$P(X = 6) = {}^6C_6 \left(\frac{1}{4}ight)^6 \left(\frac{3}{4}ight)^0 = 1 	imes \frac{1}{4^6} 	imes 1 = \frac{1}{4^6}$.Therefore,$P(X \ge 5) = \frac{18}{4^6} + \frac{1}{4^6} = \frac{19}{4^6} = \frac{19}{4096}$.

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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