Candidates’ performance in this question was fair. They performed better in the part (b)
than in the part (a).
In part (a), ncr + ncr-1 = n! + n! = n!(n-r+1)+n!(r)
r!(n-r) (r-1)!(n-r+1)! r!(n-r+1)!
= n!(n+1) = (n+1)! = n+1
Cr
r!(n-r+1)! r!(n+1-r)
In part (b) probability of 2 obtaining distinction = 6
C2
(0.04)2
(0.96)4
= 0.0204.
Probability of at most 3 of them obtaining distinction = 1-p(more than three candidates obtaining distinction) = 1- (6
C4
(0.04)4
(0.96)2
+ 6
C5
(0.04)5 (0.96) + 6
C6
(0.04)6
) = 1 – (0.000035389 + 0.000000589 + 0.000000004) = 0.99996 approximately.