A problem in Statistics is given to four stud | vedclass

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(B) Let $P(A), P(B), P(C),$ and $P(D)$ be the probabilities of students $A, B, C,$ and $D$ solving the problem respectively.Given: $P(A) = \frac{1}{3}

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$\frac{2}{3}$

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(B) Let $P(A), P(B), P(C),$ and $P(D)$ be the probabilities of students $A, B, C,$ and $D$ solving the problem respectively.Given: $P(A) = \frac{1}{3}, P(B) = \frac{1}{4}, P(C) = \frac{1}{5}, P(D) = \frac{1}{6}$.The probability that a student fails to solve the problem is $1 - P(	ext{solving})$.$P(A') = 1 - \frac{1}{3} = \frac{2}{3}$$P(B') = 1 - \frac{1}{4} = \frac{3}{4}$$P(C') = 1 - \frac{1}{5} = \frac{4}{5}$$P(D') = 1 - \frac{1}{6} = \frac{5}{6}$Since the events are independent,the probability that none of them solve the problem is $P(A') 	imes P(B') 	imes P(C') 	imes P(D') = \frac{2}{3} 	imes \frac{3}{4} 	imes \frac{4}{5} 	imes \frac{5}{6} = \frac{2}{6} = \frac{1}{3}$.The probability that the problem will be solved is $1 - P(	ext{none solve}) = 1 - \frac{1}{3} = \frac{2}{3}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

$A$ problem in Statistics is given to four students $A, B, C$ and $D$. Their chances of solving it are $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ and $\frac{1}{6}$ respectively. What is the probability that the problem will be solved?

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