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, then the probabilities that the problem is solved and exactly one of them solve the problem respectively are

• A. $\frac{1}{3}$, $\frac{2}{3}$
• B. $\frac{2}{3}$, $\frac{1}{2}$
• C. $\frac{1}{5}$, $\frac{1}{3}$
• D. $\frac{1}{2}$, $\frac{2}{3}$

Answer 1

Answer: (B)

Probability of solving the problem by $A$ i.e.
$P(A) = \frac{1}{2}$
Probability of solving the problem by $B$, i.e.
$P(B) = \frac{1}{3}$
Probability of not solving the problem by $A$
$= P\left(A'\right) = 1-P \left(A \right) = 1 -\frac{1}{2}=\frac{1}{2}$
Probability of not solving the problem by $B$
$= P\left(B'\right) = 1-P \left(B \right) = 1 -\frac{1}{3} = \frac{2}{3}$
(i) $P$(the problem is solved)
$= 1 - P$(none of them solve the problem)
$= 1-P\left(A' \cap B'\right) = 1-P\left(A'\right) P\left(B'\right)$
($\because A$ and $B$ are independent $\Rightarrow A'$ and $B'$ are also independent)
$= 1-\left(\frac{1}{2}\times \frac{2}{3}\right) = 1-\frac{1}{3}=\frac{2}{3}$
(ii) $P$(exactly one of them solve the problem)
$= P\left(A\right) P\left(B'\right) + P\left(A'\right) P\left(B\right)$
$= \frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}$
$= \frac{1}{3}+\frac{1}{6} = \frac{1}{2}$