P(A) = \(\frac{3}{5}\), P(B) = \(\frac{2}{5}\), P(C) = \(\frac{3}{4}\). In order that two shots may hit the target the following three mutually exclusive events are possible,
(i) A and B hit the target and not C.
(ii) B and C hit the target and not A.
(iii) A and C hit the target and not B.
If p1, p2, p3 are the probabilities of these three events then
p1 = P (A). P (B) . P (not C) = \(\frac{3}{5}\) x \(\frac{2}{5}\) x \(\big(1-\frac{3}{4}\big)\) = \(\frac{3}{5}\) x \(\frac{2}{5}\) x \(\frac{1}{4}\) = \(\frac{6}{100}\)
p2 = P (B). P (C) . P (not A) = \(\frac{2}{5}\) x \(\frac{3}{4}\) x \(\big(1-\frac{3}{5}\big)\) = \(\frac{2}{5}\) x \(\frac{3}{4}\) x \(\frac{2}{5}\) = \(\frac{12}{100}\)
p3 = P (A) × P (C) × P (not B) = \(\frac{3}{5}\) x \(\frac{3}{4}\) x \(\big(1-\frac{2}{5}\big)\) = \(\frac{3}{5}\)x \(\frac{3}{4}\) x \(\frac{3}{5}\) = \(\frac{27}{100}\)
∴ Required probability = p1 + p2 + p3 = \(\frac{6}{100}\) + \(\frac{12}{100}\) + \(\frac{27}{100}\) = \(\frac{45}{100}\) = \(\frac{9}{20}.\)