(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%
