Further Mathematics Paper 2, May/June. 2015

Question 13

  1. Edem and his wife were invited to a dinner by a family of 5. They all sat in such a way that Edem sat next to his wife. Find the number of ways of sitting them in a row.
  2. A bag contains 4 red and 5 black identical balls. If 5 balls are selected at random from the bag, one after the other with replacement, find the probability that:

(i) a red ball was picked 3 times;

(ii) a black ball was picked at most 2 times.



The Chief Examiner reported that this question attracted majority of the candidates and described their performance as fair. According to the report, majority of the candidates performed better in part (b) than in part (a). In part (a), candidates were reported to have performed poorly. Candidates were expected to recognize the problem as that of permutation. There were two ways of sitting Edem and his wife. Taking them as one person, there would then be 6! Ways of sitting everybody at the table. Therefore, the number of ways of sitting at the table if Edem is to sit next to his wife = 2 × 6! Ways = 1440 ways. In part (b), candidates performed better. They were reported to recognize that the problem followed a binomial distribution, P(x = r) = nCrprqn – r, where n = number of trials, p = probability of success, q = probability of failure and r = favourable number of times. In part (b)(i), n = 5, r = 3, p =  and q = . Therefore, P(x = 3) = 5C3 = 10 = 0.270961. In part (b)(ii), n = 5, r = 0 or 1 or 2, p =  and q =  . Therefore, the required probability was P(x = 0) + P(x = 1) + P(x = 2) = 5C0 + 5C1 + 5C2 = 0.396687.