Question

The probabilities of solving a particular problem by $A$ and $B$ independently are respectively $\tfrac{1}{2}$ and $\tfrac{1}{3}$. If both try to solve the problem independently, find the probability that (i) the problem is solved, (ii) only one of them solves the problem.

Hint:

For "at least one solves," use $1 - P(\text{none})$. For "only one," add the probabilities of each solving while the other fails.

Answer 1

Let $P(A) = \tfrac{1}{2}$ be the probability that $A$ solves the problem. Let $P(B) = \tfrac{1}{3}$ be the probability that $B$ solves the problem. Thus, \[ P(A') = 1 - P(A) = \tfrac{1}{2}, P(B') = 1 - P(B) = \tfrac{2}{3}. \] (i) Probability that the problem is solved. The problem is solved if either $A$ solves, or $B$ solves, or both solve. \[ P(\text{problem solved}) = 1 - P(\text{none solves}) \] \[ = 1 - P(A' \cap B') = 1 - \left(\tfrac{1}{2} \times \tfrac{2}{3}\right) = 1 - \tfrac{1}{3} = \tfrac{2}{3}. \] (ii) Probability that only one of them solves the problem. Case 1: $A$ solves, $B$ does not: \[ P(A \cap B') = \tfrac{1}{2} \times \tfrac{2}{3} = \tfrac{1}{3}. \] Case 2: $B$ solves, $A$ does not: \[ P(A' \cap B) = \tfrac{1}{2} \times \tfrac{1}{3} = \tfrac{1}{6}. \] So, \[ P(\text{only one solves}) = \tfrac{1}{3} + \tfrac{1}{6} = \tfrac{1}{2}. \] Final Answer: \[ (i)\; \tfrac{2}{3}, (ii)\; \tfrac{1}{2} \]