This was also another question where candidates' performance was reported to be commendable.
In part (a), candidates were able to show that the number of ways of selecting 3 students from 9 students = 9C3 = 84 ways. They also showed that probability of selecting at least 2 girls = prob[(2 girls and one boy) or (3 girls and no boy)]
=⁴C₂X⁵C₁ + ⁴c₃X⁵C_o = 30 +4 = 34 = 0.405.
                    C3               C3              84      84

In part (b), candidates were able to recognize that the problem followed a binomial probability distribution with n = 5, p = 0.8 and q = 1- 0.8 = 0.2.
Prob(no goal) ⁵c₀ (0.8)⁰(0.2)⁵ = 0.0003.
P(at most @ goals) ⁵c₀ (0.8)⁰(0.2)⁵ + ⁵c₁(0.8)¹(0.2)⁴ + ⁵c₂(0.8)²(0.2)³

= 0.00032 + 0.0064 + 0.0512 = 0.0579