Probability that A solves the problem = \(\frac{1}{2}\)
⇒ Probability that A does not solve the problem P(\(\overline{\mathrm{A}}\))
= 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\).
Probability that B solves the problem = \(\frac{1}{3}\)
⇒ Probability that B does not solve the problem
= 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\).
(i) Probability that problem is not solved P(\(\overline{\mathrm{A}}\) \(\overline{\mathrm{B}}\))
= P(\(\overline{\mathrm{A}}\)) P(\(\overline{\mathrm{B}}\)) = \(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\).
⇒ Probability that problem is solved
= 1 – P(\(\overline{\mathrm{A}}\) \(\overline{\mathrm{B}}\)) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\).
(ii) Exactly one of them solves the problem
= P(A \(\overline{\mathrm{B}}\)) + P(\(\overline{\mathrm{A}}\) B)
= P(A) P(\(\overline{\mathrm{B}}\)) + P(\(\overline{\mathrm{A}}\)) P(B)
Since A and B are independent events.
So, A \(\overline{\mathrm{B}}\) and \(\overline{\mathrm{A}}\) B are also independent.
Now, P(A) = \(\frac{1}{2}\), P(\(\overline{\mathrm{B}}\)) = \(\frac{2}{3}\),
P(\(\overline{\mathrm{A}}\)) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\)
∴ Probability that exactly one of them solves the problem
= \(\frac{1}{2}\) × \(\frac{2}{3}\) + \(\frac{1}{2}\) × \(\frac{1}{3}\) = \(\frac{1}{3}\) + \(\frac{1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\).
