The probabilities that Ali, Baba and Katty will gain admission to college are \(\frac{2}{3}, \frac{3}{4}\) and \(\frac{4}{5}\) respectively. Find the probability that:
(a) only Katty and Baba will gain admission ;
(b) none of them will gain admission ;
(c) at most two of them will gain admission.
\(p(Ali) = p(A) = \frac{2}{3} ; p(Baba) = p(B) = \frac{3}{4} ; p(Katty) = p(K) = \frac{4}{5}\)
(a) \(p(\text{only Katty and Baba}) = p(K) \times p(B) \times p(A')\)
= \(\frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}\)
= \(\frac{1}{5}\)
(b) \(p(\text{none gains admission}) = p(A') \times p(K') \times p(B')\)
= \(\frac{1}{3} \times \frac{1}{4} \times \frac{1}{5}\)
= \(\frac{1}{60}\).
(c) \(p(\text{at most two will gain admission}) = p(\text{none gains}) + p(\text{one gains}) + p(\text{two gains})\)
\(p(\text{one gains}) = p(A) \times p(B') \times p(K') + p(B) \times p(A') \times p(K') + p(K) \times p(A') \times p(B')\)
= \(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5} + \frac{3}{4} \times \frac{1}{3} \times \frac{1}{5} + \frac{4}{5} \times \frac{1}{3} \times \frac{1}{4}\)
= \(\frac{2}{60} + \frac{3}{60} + \frac{4}{60}\)
= \(\frac{9}{60}\)
\(p(\text{two gains}) = p(A) \times p(B) \times p(K') + p(A) \times p(K) \times p(B') + p(B) \times p(K) \times p(A')\)
= \(\frac{2}{3} \times \frac{3}{4} \times \frac{1}{5} + \frac{2}{3} \times \frac{4}{5} \times \frac{1}{4} + \frac{3}{4} \times \frac{4}{5} \times \frac{1}{3}\)
= \(\frac{6}{60} + \frac{8}{60} + \frac{12}{60}\)
= \(\frac{26}{60}\)
\(p(\text{at most two gains admission}) = \frac{1}{60} + \frac{9}{60} + \frac{26}{60}\)
= \(\frac{36}{60}\)
= \(\frac{3}{5}\).
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