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 - [^{5}C_{a} (1/6)^{O} (5/6)^{5} + 5C_{1} (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 ^{n}C_{r} =* n!*

r!(n-r)!

and ^{n}P_{r} =* n!*

r!(n-r)!

Therefore, ^{x}c_{3}/^{x}p_{2}= x!/3!(x-3)!/x!/(x-2)! = x!/3!(x-3)! × (x-2)!/x! = (x-2/3!)

Since ^{x}c_{3}/^{x}p_{2}= 2/3 ⇒ x-2/3! = 2/3 i.e 3x - 6 = 12 ⇒ x = 6. * *