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.
Explanation
(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}).