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 5

A committee of four persons is to be formed from 7 girls and 5 boys. Calculate, correct to two decimal places, the probability that the committee will consist of:
(a) 2 boys and 2 girls;
(b) at least 3 boys.

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Observation

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.