Q78 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
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
- A$\frac{3}{4}$
- Correct answerB$\frac{11}{24}$
- C$\frac{23}{24}$
- D$\frac{1}{4}$
Solution
The required probability
$$
\begin{aligned}
=P(A) & P(\bar{B}) P(\bar{C})+P(\bar{A}) P(B) P(\bar{C})+P(\bar{A}) P(\bar{B}) P(C) \\
& =\left(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4}\right) \\
& =\frac{1}{4}+\frac{1}{8}+\frac{1}{12}=\frac{11}{24}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.