Question

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

• A. $\frac{25}{52}$
• B. $\frac{25}{56}$
• C. $\frac{13}{42}$
• D. $\frac{11}{45}$

Answer 1

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

We have

$P\left(A\right)=\frac{1}{3},P\left(A^{\prime }\right)=\frac{2}{3},$

$P\left(B\right)=\frac{2}{7},P\left(B^{\prime }\right)=\frac{5}{7},$

$P\left(C\right)=\frac{3}{8},P\left(C^{\prime }\right)=\frac{5}{8}$

Probability that only one will solve the problem

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

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

$=\frac{25}{168}+\frac{20}{168}+\frac{30}{168}=\frac{25}{56}$