Question

The probability of A, B, and C solving a problem are $\frac{1}{2},\frac{2}{7},\frac{3}{8}$ respectively. If all the three try to solve the problem simultaneously, the probability that exactly one of them will solve it is

• A. $\frac{25}{168}$
• B. $\frac{25}{56}$
• C. $\frac{20}{168}$
• D. $\frac{30}{168}$

Answer 1

The correct option is B

$\frac{25}{56}$

Explanation for the correct option

Let $P\left(A\right)$ be the probability that A solves the problem $=\frac{1}{3}$

$P\left(B\right)$ be the probability that B solves the problem$=\frac{2}{7}$

$P\left(C\right)$ be the probability that C solves the problem $=\frac{3}{8}$

$$\begin{gathered} P(A\text{'})=1-\frac{1}{3}=\frac{2}{3} \\ P(B\text{'})=1-\frac{2}{7}=\frac{5}{7} \\ P(C\text{'})=1-\frac{3}{8}=\frac{5}{8} \end{gathered}$$

The probability that exactly one of them will solve the problem $=$ Probability that exactly A solves the problem $+$ probability that exactly B solves the problem $+$ probability that exactly C solves the problem

$$\begin{gathered} =P\left(A\right)P(B\text{'})P(C\text{'})+P(A\text{'})P\left(B\right)P(C\text{'})+P(A\text{'})P(B\text{'})P\left(C\right) \\ =\left(\frac{1}{3}\right)\left(\frac{5}{7}\right)\left(\frac{5}{8}\right)+\left(\frac{2}{3}\right)\left(\frac{2}{7}\right)\left(\frac{5}{8}\right)+\left(\frac{2}{3}\right)\left(\frac{5}{7}\right)\left(\frac{3}{8}\right) \\ =\left(\frac{25}{168}\right)+\left(\frac{20}{168}\right)+\left(\frac{30}{168}\right) \\ =\frac{75}{168} \\ =\frac{25}{56} \end{gathered}$$

Hence the correct option is option(B) i.e. $\frac{25}{56}$.