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Further Mathematics Paper 2, Nov/Dec. 2008  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
General Comments
Weakness/Remedies
Strength
Question 15

(a) Simplify   nCr   +  nCr-1
(b) In an examination 4% of the candidates passed with distinction.  If 6 of the candidates are selected at random, what is the probability that:

  • 2 of them obtained distinction.
  • at most 3 of them obtained distinction
_____________________________________________________________________________________________________
Observation

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+1Cr
                                      r!(n-r+1)!                   r!(n+1-r)

In part (b) probability of 2 obtaining distinction = 6C2 (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- (6C4 (0.04)4 (0.96)2 + 6C5 (0.04)5 (0.96) + 6C6 (0.04)6) =  1 – (0.000035389 + 0.000000589 + 0.000000004)  = 0.99996 approximately.

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