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 + *C**s**(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__ •