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A problem is given to 3 students $A, B$ and $C$ whose chances of solving it are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. Then, the probability of the problem being solved by exactly one of them, if all the three try independently, is
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Verified Answer
The correct answer is:
$\frac{11}{24}$
The required probability
$$
\begin{aligned}
=P(A) & P(\bar{B}) P(\bar{C})+P(\bar{A}) P(B) P(\bar{C})+P(\bar{A}) P(\bar{B}) P(C) \\
& =\left(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4}\right) \\
& =\frac{1}{4}+\frac{1}{8}+\frac{1}{12}=\frac{11}{24}
\end{aligned}
$$
$$
\begin{aligned}
=P(A) & P(\bar{B}) P(\bar{C})+P(\bar{A}) P(B) P(\bar{C})+P(\bar{A}) P(\bar{B}) P(C) \\
& =\left(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4}\right) \\
& =\frac{1}{4}+\frac{1}{8}+\frac{1}{12}=\frac{11}{24}
\end{aligned}
$$
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