Further Mathematics Paper 2, May/June. 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 14

(a) Eight coins are tossed once. Find, correct to three decimal places, the probability of obtaining
(i) exactly 8 heads;
(ii) at least 5 heads;
(iii) at most 1 head.

(b) In how many ways can four letters from the word SHE E P be arranged.
(i) without any restriction
(ii) with only one E.

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Observation

The chief examiner reported that this question was poorly handled by majority of the candidates. Application of binomial expansion theorem to solving problems in statistics needs to be emphasized by teachers during
instruction.
In part (a), candidates were expected to apply the formula nCr pr qn-r where n = total number of outcomes, r = number of required outcomes, p = probability of success and q = probability of failure. Here, n = 8, P = q = Yz. For the probability of obtaining exactly 8 heads, r = 8. Therefore the probability was 8Cge/2)8 eh)o= 0.003906 == 0.004. Probability of obtaining at least 5 heads was given by 8Cs e/2ie/2)3 + 8C6 (lh)6(1/2l + SC7(1/2)\1/2)1 +sCs e/2)s(I/2)O = 0.3297 = 0.330, correct to 3 decimal places. Probability of at most 1 head = SCo (1/2)°(1/2)8 + SC1C1h)leiz/ = 0.035156 == 0.035.

In part (b), the number of ways of picking 4 letters from 5 letters = Sp 4. Since E appeared 2 times, then number of way of picking the 4 letters became~ = 120
2        2 = 60 ways.
If the four letters is to contain one E, then the E could be picked in 2 ways. Thus, taking the 2 Es as one, the probability of picking 4 letters becomes 2 x 4p 4 = 2 x 24 = 48 ways.