This question was reported to have been attempted by majority of the candidates and they
performed well in it. However, some of them mistook it for a problem on Binomial distribution
while others took it for a problem on permutation.

Candidates were expected to recognize it as a problem on combination (selection of objects in
no particular order).

In part (a), candidates' expected responses were as follows:

Total number of persons = 12. Number of ways of selecting 4 persons from 12 persons is given
by ^{12}C_{4} = 12!/4!8! = 495 ways. Number of ways of selecting 2 girls and 2 boys from 7 girls and 5
boys respectively = ^{7}C_{2} x^{5}C_{2} = 21 x 10 = 210 ways. Therefore, probability of selecting 2 girls
and 2 boys = 210/495= 0.42 to 2 decimal places.

In part (b), selecting at least 3 boys implied selecting 3 boys or 4 boys. This can be done in

^{7}C_{1} x ^{5}C_{3}x^{7}C_{0} X ^{5}C_{4} = 70 + 5 = 75 ways. Therefore, probability of selecting at least 3 boys = 75/495

= 0.15.