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)]

=^{4}C_{2}X^{5}C_{1} + ^{4}c_{3}X^{5}C_{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) ^{5}c_{0} (0.8)^{0}(0.2)^{5} = 0.0003.

P(at most @ goals) ^{5}c_{0} (0.8)^{0}(0.2)^{5} + ^{5}c_{1}(0.8)¹(0.2)^{4} + ^{5}c_{2}(0.8)²(0.2)³

= 0.00032 + 0.0064 + 0.0512 = 0.0579 * *