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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 problem
$\frac{1}{2}$
$\frac{1}{2}$
$\frac{1}{2}$
$\frac{1}{2}$
Solution
Probability of solving the problem by $\mathrm{A},\, \mathrm{P}(\mathrm{A})=\frac{1}{2}$
Probability of solving the problem by $\mathrm{B}, \,\mathrm{P}(\mathrm{B})=\frac{1}{3}$
since the problem is solved independently by $A$ and $B$,
$\therefore $ $\mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$
$P(A^{\prime})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}$
$P(B^{\prime})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}$
Probability that exactly one of them solves the problem is given by,
$\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{A})$
$=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}$
$=\frac{1}{3}+\frac{1}{6}$
$=\frac{1}{2}$
