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 ¹²C₄ = 12!/4!8! = 495 ways. Number of ways of selecting 2 girls and 2 boys from 7 girls and 5 boys respectively = ⁷C₂ x⁵C₂ = 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
⁷C₁ x ⁵C₃x⁷C₀ X ⁵C₄ = 70 + 5 = 75 ways. Therefore, probability of selecting at least 3 boys = 75/495
= 0.15.