A problem in statistics is given to three stu | vedclass

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(C) Let $E_1, E_2, E_3$ be the events that students $P, Q, R$ solve the problem respectively.Given probabilities are $P(E_1) = \frac{1}{2}$,$P(E_2) =

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$\frac{3}{4}$

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(C) Let $E_1, E_2, E_3$ be the events that students $P, Q, R$ solve the problem respectively.Given probabilities are $P(E_1) = \frac{1}{2}$,$P(E_2) = \frac{1}{3}$,$P(E_3) = \frac{1}{4}$.The probability that the problem is not solved by any of them is the probability that all three fail to solve it.Since they try independently,the probability that none of them solve the problem is:$P(	ext{None solve}) = P(E_1^c) 	imes P(E_2^c) 	imes P(E_3^c)$$P(E_1^c) = 1 - \frac{1}{2} = \frac{1}{2}$$P(E_2^c) = 1 - \frac{1}{3} = \frac{2}{3}$$P(E_3^c) = 1 - \frac{1}{4} = \frac{3}{4}$$P(	ext{None solve}) = \frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4} = \frac{6}{24} = \frac{1}{4}$The probability that the problem is solved is $1 - P(	ext{None solve})$.$P(	ext{Solved}) = 1 - \frac{1}{4} = \frac{3}{4}$.

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

Expenditure in $Rs. (x)$	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
No. of families $(f)$	$24$	$33$	$37$	$b$	$25$

$A$ problem in statistics is given to three students $P, Q$ and $R$. Their chances of solving the problem are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. If all of them try independently,then the probability that the problem is solved is:

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