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 + 4c3X5Co = 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