(a) The probabilities that three boys pass an examination are (frac{2}{3}, frac{5}{8}) and (frac{3}{4}) respectively. Find the probability that :
(i) all three boys pass ; (ii) none of the boys pass ; (iii) only two of the boys pass.
(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.
Explanation
(a)(i) P(A) = (frac{2}{3}) ; P(B) = (frac{5}{8}) ; P(C) = (frac{3}{4}).
(P(A cap B cap C) = P(A) times P(B) times P(C))
= (frac{2}{3} times frac{5}{8} times frac{3}{4} = frac{5}{16})
(ii) P(A’) = (1 – frac{2}{3} = frac{1}{3})
P(B’) = (1 – frac{5}{8} = frac{3}{8})
P(C’) = (1 – frac{3}{4} = frac{1}{4})
(P(A’ cap B’ cap C’) = P(A’) times P(B’) times P(C’))
= (frac{1}{3} times frac{3}{8} times frac{1}{4} )
= (frac{1}{32})
(iii) P(A and B pass, C fails) = (frac{2}{3} times frac{5}{8} times frac{1}{4} = frac{5}{48})
P( A and C pass, B fails) = (frac{2}{3} times frac{3}{4} times frac{3}{8} = frac{3}{16})
P(B and C pass, A fails) = (frac{5}{8} times frac{3}{4} times frac{1}{3} = frac{5}{32})
P(only two of the boys pass) = (frac{5}{48} + frac{3}{16} + frac{5}{32} = frac{43}{96})
(b) Let the cost price be x
% gain = (frac{text{actual gain}}{text{cost price}} times 100%)
(20 = frac{36,000 – x}{x} times 100%)
(frac{36,000 – x}{x} = frac{1}{5})
(5(36,000 – x) = x implies 180,000 – 5x = x)
(180,000 = x + 5x = 6x)
(x = N30,000)
Discount (5%) : (frac{5}{100} times N36,000 = N1,800)
Selling price : (N(36,000 – 1,800) = N34,200)
Actual gain : (frac{34,200 – 30,000}{30,000} times 100% )
= 14%