A problem is given to 3 students $A, B$ and $C$ whose chances of solving it are $\frac{1}{2}, \frac{1}{3}$…

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
  1. $\frac{3}{4}$
  2. $\frac{11}{24}$
  3. $\frac{23}{24}$
  4. $\frac{1}{4}$

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} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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