Further Mathematics Paper 2, Nov/Dec. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 15

((a) A team of 3 students is to be selected from 5 boys and 4 girls to represent a school in a quiz.
(i) If there is no restriction, in how many ways can the team be selected?
(ii) Calculate, correct to three decimal places, the probability that there will be at least two girls in the team.
(b) The probability that a player will score a penalty kick is 0.8. If he takes 5 penalty kicks, calculate, correct to four decimal places, the probability that he will score:
(i) no goal;
(ii) at most two goals.

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Observation

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