Probability of solving specific problem independently by $A$ and $B$ are $\cfrac{1}{2}$ and $\cfrac{1}{3}$ respectively. If both try to solve the problem independently, then find the probability that
the problem is solved

(ii) exactly one of them solves the problem.

.: Let ${E_1}:$ $A$solves the problem’ and ${E_2}$ : ‘ $B$ solves the problem’,

then $P\left( {{E_1}} \right) = \cfrac{1}{2}$

and $P\left( {{E_2}} \right) = \cfrac{1}{3}$
.
(i) $P$ (the problem is solved)

Probability that the problem is solved by the least one of

$A$ and $B = P\left( {{E_1}} \right)P\left( {{{\bar E}_2}} \right) + P\left( {{{\bar E}_1}} \right)P\left( {{E_2}} \right) + P\left( {{E_1}} \right)P\left( {{E_2}} \right)$

$= \left( {\cfrac{1}{2} \times \cfrac{2}{3}} \right) + \left( {\cfrac{1}{2} \times \cfrac{1}{3}} \right) + \cfrac{1}{2} \times \cfrac{1}{3} = \cfrac{1}{3} + \cfrac{1}{6} + \cfrac{1}{6} = \cfrac{{8 + 4 + 4}}{{24}} = \cfrac{{16}}{{24}} = \cfrac{2}{3}$

(ii) $P$ ( Exactly one of them solves the problem)

$= P$ ( $A$ solves the problem and $B$ does not) $+ P$ ( $B$ solves the problem and $A$ does not)

$= P\left( {{E_1} \cap {{\bar E}_2}} \right) + P\left( {{E_2}\frown {{\bar E}_1}} \right)$

$= P\left( {{E_1}} \right)P\left( {{{\bar E}_2}} \right) + P\left( {{E_2}} \right)P\left( {{{\bar E}_1}} \right)$

( ${E_1}$ \& ${E_2}$ are independent)
$= P\left( {{E_1}} \right)\left( {1 - P\left( {{E_2}} \right)} \right) + P\left( {{E_2}} \right)\left( {1 - P\left( {{E_1}} \right)} \right)$

$= \cfrac{1}{2}\left( {1 - \cfrac{1}{3}} \right) + \cfrac{1}{3}\left( {1 - \cfrac{1}{2}} \right) = \cfrac{1}{2} \times \cfrac{2}{3} + \cfrac{1}{3} \times \cfrac{1}{2} = \cfrac{1}{3} + \cfrac{1}{6} = \cfrac{1}{2}$