Question
Probability of solving specific problem independently by A and B are $\frac{1}{2}\ \text{and}\ \frac{1}{3}$ respectively.

If both try to solve the problem independently, find the probability that

  1. The problem is solved.
  2. Exactly one of them solves the problem.

Answer

Probability of solving the problem by A, P(A) $=\frac{1}{2}$

Probability of solving the problem by B, P(B) $=\frac{1}{3}$

Since the problem is solved independently by A and B,

$\therefore\text{P}(\text{AB})=\text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$

$\text{P}(\text{A}')=1-\text{P}(\text{A})=1-\frac{1}{2}=\frac{1}{2}$

$\text{P}(\text{B}')=1-\text{P}(\text{B})=1-\frac{1}{3}=\frac{2}{3}$

  1. Probability that the problem is solved $=\text{P}(\text{A}\cup\text{B})$

$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{AB})$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$=\frac{4}{6}$

$=\frac{2}{3}$

  1. Probability that exactly one of them solves the problem is given by,

$\text{P}(\text{A}).\text{P}(\text{B}')+\text{P}(\text{B}).\text{P}(\text{A}')$

$ =\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3}$

$=\frac{1}{3}+\frac{1}{6}$

$=\frac{1}{2}$

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