Given,
$P(A)=$ Probability of solving the problem by $A=1 / 2$
Concept: $\mathrm{P}(\mathrm{B})=$ Probability of solving the problem by $\mathrm{B}=1 / 3$
Because $A$ and $B$ both are independent.Two events are independent, statistically independent, or stochastically independent if the occurrence of one does not affect the probability of occurrence of the other.
$\Rightarrow P(A \cap B)=P(A) . P(B)$
$\Rightarrow P(A \cap B)=1 / 2 \times 1 / 3=1 / 6$
$P\left(A^{\prime}\right)=1-P(A)=1-1 / 2=1 / 2$
$P\left(B^{\prime}\right)=1-P(B)=1-1 / 3=2 / 3$
(i) The problem is solved
The problem has been solved, i.e. it has been solved by either A or B.
$=P(A \cup B)$
As we know, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=1 / 2+1 / 3-1 / 6=4 / 6$
$\Rightarrow P(A \cup B)=2 / 3$
(ii) Exactly one of them solves the problem
That is either problem is solved by A but not by B or vice versa
That is $\mathrm{P}$ (A).P (B $\left.^{\prime}\right)+\mathrm{P}$ (A’).P (B)
$=1 / 2(2 / 3)+1 / 2(1 / 3)$
$=1 / 3+1 / 6=3 / 6$ $\Rightarrow P(A) \cdot P\left(B^{\prime}\right)+P\left(A^{\prime}\right) . P(B)=1 / 2$