Probability of solving specific problem independently by \mathrm{A} and \mathrm{B} are \mathbf{1} / \mathbf{2} and \mathbf{1} / \mathbf{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.
Probability of solving specific problem independently by \mathrm{A} and \mathrm{B} are \mathbf{1} / \mathbf{2} and \mathbf{1} / \mathbf{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.

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