Question

Probability of solving specific 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
(i) the problem is solved
(ii) exactly one of them solves the problem.

• A. $0.67,0.78$
• B. $0.66,0.50$
• C. $0.67,0.98$
• D. $0.66,0.44$

Answer 1

Answer: (B)

$0.66,0.50$

Probability of solving the problem by $A$, $P\left(A\right)=\frac{1}{2}$
Probability of solving the problem by $B$, $P\left(B\right)=\frac{1}{3}$
Since the problem is solved independently by $A$ and $B$,
$\therefore P\left(AB\right)=P\left(A\right)\cdot P\left(B\right)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$
$P\left(A\prime \right)=1-P\left(A\right)=1-\frac{1}{2}=\frac{1}{2}$
$P\left(B\prime \right)=1-P\left(B\right)=1-\frac{1}{3}=\frac{2}{3}$
(i) Probability that the problem is solved $=P(A\cup B)$
$=P\left(A\right)+P\left(B\right)-P\left(AB\right)$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$
$=\frac{4}{6}$
$=\frac{2}{3}=0.66$
(ii) Probability that exactly one of them solves the problem is given by,
$P\left(A\right)\cdot P\left(B\prime \right)+P\left(B\right)\cdot P\left(A\prime \right)$
$=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}$
$=\frac{1}{3}+\frac{1}{6}$
$=\frac{1}{2}=0.50$