It was reported that this question was not as popular as questions 13 and 14.  It was also reported to be poorly answered by majority of those who attempted it.  Candidates’ were expected to recognize that the probability problem followed a binomial probability distribution nCrprqr with n = 4, p =   and q = .  Probability that all lines were engaged was 4C440 =  or 0.2963.  Probability that 2 lines were engaged was 4C2 =  or 0.2963. Probability that none of the lines was engaged was 4C004 =  or 0.1975.  Probability that at least one line was engaged  =  (1 – probability that no line was engaged) i.e. 1 – (4C0)= 1 -  =  or 0.8023.