Question

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

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

Answer 1

The correct option is B $\frac{25}{56}$

Probability exactly one of A,B or C solves it

= probability of A solving (remaining fail) + probability of B solving (remaining fail) + probability of C solves (remaining fail)

$=P\left(A\right)\times P\left(B^C\right)\times P\left(C^C\right)+P\left(A^C\right)P\left(B\right)P\left(C^C\right)+P\left(A^C\right)P\left(B^C\right)P\left(C\right)$

$=\left(\frac{1}{3}\right)(1-\frac{2}{7})(1-\frac{3}{8})+(1-\frac{1}{3})\left(\frac{2}{7}\right)(1-\frac{3}{8})+(1-\frac{1}{3})(1-\frac{2}{7})\left(\frac{3}{8}\right)$

$=\frac{1}{3}\times \frac{5}{7}\times \frac{5}{8}+\frac{2}{3}\times \frac{2}{7}\times \frac{5}{8}+\frac{2}{3}\times \frac{5}{7}\times \frac{3}{8}$

$\Rightarrow \frac{25}{168}+\frac{20}{168}+\frac{30}{168}\Rightarrow \frac{75}{168}\Rightarrow \frac{25}{56}$

probability that exactly one will solve is $\frac{25}{56}$