Question
Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that exactly one of them solves the problem.
✓
Answer
Given:
P(A) = Probability of solving the problem by A = $\frac{1}{2}$
P(B) = Probability of solving the problem by B = $\frac{1}{3}$
P(A') = $1 - \frac{1}{2} = \frac12$
and P(B') = $1 - \frac{1}{3} = \frac23$
Since, A and B are independent.
Now, P (exactly one of them solves) = Either problem is solved by A but not by B or vice versa
= P(A).P(B’) + P(A’).P(B)
= $\frac{1}{2} \cdot \frac{2}{3}+\frac{1}{2} \cdot \frac{1}{3}$
= $\frac{1}{3}+\frac{1}{6}=\frac{3}{6}$
⇒ P(A).P(B') + P(A').P(B) = $\frac{1}{2}$
P(A) = Probability of solving the problem by A = $\frac{1}{2}$
P(B) = Probability of solving the problem by B = $\frac{1}{3}$
P(A') = $1 - \frac{1}{2} = \frac12$
and P(B') = $1 - \frac{1}{3} = \frac23$
Since, A and B are independent.
Now, P (exactly one of them solves) = Either problem is solved by A but not by B or vice versa
= P(A).P(B’) + P(A’).P(B)
= $\frac{1}{2} \cdot \frac{2}{3}+\frac{1}{2} \cdot \frac{1}{3}$
= $\frac{1}{3}+\frac{1}{6}=\frac{3}{6}$
⇒ P(A).P(B') + P(A').P(B) = $\frac{1}{2}$
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