Candidates were expected to apply the binomial rule nCrprqn-r, (where p = probability of success, q = probability of failure, n = 5, r = number of required outcome ),to solve this problem.

In part (a) n = 5, r = 3, P = ~ = 0.6 q = ~ = 0.4. The probability of selecting 3 boys
                                                       5                     5
         5              3           2       5        3 3 2 2
was ClO.6) (0.4) or C3 (/5) (/5) = 0.3456 or 625·
In part (b), candidates were expected to show that the probability of selecting at least 3 girls implied probability of selecting 3 girls, P(3) or probability of selecting 4 girls, P( 4), or probability of selecting 5 girls, P(5). This could be obtained in two ways; either P(3) + P(4) + P~5) or 1- [P(O) + P(1) + P(2)]
i.e. either 5C3(0.4i(0.5t + 5C4(0.4)4(0.6)J + Cs(0.4i(0.6)o or
1 - [SCo (0.4)°(0.6)5 + C1 (0.4)1(0.6)4 + 5C2(0.4)\0.6)4], which after a littleComputation gave the probability of at least 3 girls as 0.3174 or 39:225 •