Question

Probabilities of A,B and C of solving a problem are $\frac{1}{3},\frac{1}{2}and\frac{1}{4}$ respectively. If they all try to solve the problem then find the probability that exactly one of them will solve the problem.

Answer 1

Given:-

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

$P\left(B\right)=\frac{1}{2}\Rightarrow P\left(B^{\prime }\right)=1-\frac{1}{2}=\frac{1}{2}$

$P\left(C\right)=\frac{1}{4}\Rightarrow P\left(C^{\prime }\right)=1-\frac{1}{4}=\frac{3}{4}$

Since $A,B$ and $C$ are independent.

Therefore,

Probability of exactly one of them will solve the problem $=P(A\cap B^{\prime }\cap C^{\prime })+P(A^{\prime }\cap B\cap C^{\prime })+P(A^{\prime }\cap B^{\prime }\cap C)=P\left(A\right)\cdot P\left(B^{\prime }\right)\cdot P\left(C^{\prime }\right)+P\left(A^{\prime }\right)\cdot P\left(B\right)\cdot P\left(C^{\prime }\right)+P\left(A^{\prime }\right)\cdot P\left(B^{\prime }\right)\cdot P\left(C\right)=\frac{1}{3}\times \frac{1}{2}\times \frac{3}{4}+\frac{2}{3}\times \frac{1}{2}\times \frac{3}{4}+\frac{2}{3}\times \frac{1}{2}\times \frac{1}{4}=\frac{3+6+2}{24}=\frac{11}{24}$

Hence the probability that exactly one of them will solve the problem is $\frac{11}{24}$.