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 14

(a) A box P contains 3 white and 5 green identical balls. Another box Q contains 6 white and 4 green identical balls.
A ball is drawn at random from P and dropped into Q. A ball is then drawn at random from Q. Find the probability that the ball drawn from Q is green.
(b) A fair die is thrown five times.
Find, correct to three decimal places, the probability of obtaining at least two sixes.
(c) If xC3        = 2/3, find x.
xP2

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Observation

It was reported that majority of the candidates did not attempt this question and the few who did performed poorly. Teachers are therefore encouraged to emphasize this area 'of the syllabus during instruction.
In part (a), candidates were expected to show that picking a green ball from Q implied (picking a green ball from P and a green ball from Q) or (not picking a green ball from P and picking a green ball from Q).
Probability of picking a green ball from P = 5/8. Probability of not picking a green ball from P = 3/8. If a ball is added to Q, the total number of balls in Q will then become 11.
Therefore, probability that a ball drawn from Q is green is given by:
P(ball added IS green or ball added IS not green) = (5/8 x 5/11) + (3/8 x 4/11) = (25/88 + 12/88) = 37/88 or 0.42.
In part (b),candidates were expected to recognize that the probability follows a binomial distribution with n = 5, p = 1/8 and q = 5/6. The probability of obtaining at least two sixes =1 - [5Ca (1/6)O (5/6)5 + 5C1 (1/6)¹(5/6)4] = 0.196 .
The report stated that this was the most popular of the three parts and they performed better than they did in the other two parts.
In part (c), candidates were expected to recall that nCr = n!
r!(n-r)!

and nPr = n!
r!(n-r)!

Therefore, xc3/xp2= x!/3!(x-3)!/x!/(x-2)! = x!/3!(x-3)! × (x-2)!/x! = (x-2/3!)

Since xc3/xp2= 2/3 ⇒ x-2/3! = 2/3 i.e 3x - 6 = 12 ⇒ x = 6.