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Further Mathematics Paper 2, Nov/Dec. 2009  
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 6

The chances of three hunters hitting a target are 1/3, 2/3 and 1/5 respectively.  If they fired independently at the target, find the probability that

  1. at least one of them hit the target.
  2. only two of them hit the target.

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Observation

Only very few candidates reportedly performed well in this question.  Majority of them displayed poor knowledge of the probability of independent events.  They were expected to show that if P(H1) = 1/3, P(H2) = 2/3 and P(H3) = 1/5 are the probabilities of each hunter hitting the target, then the probability of none of them hitting the target = P(H’1) x P(H’2) x  P(H’3) = 2/3 x 1/3 x 4/5 = 8/45, where p(H1’), p(H2’) and p(H3’) are the probabilities of each hunter not hitting the target respectively.  Therefore probability of at least one of them hit = 1 – prob. that none of them hit = 1 – 8/45 = 37/45.
Probability of only two of them hitting the target = P(H1 H2 H’3 + H’1 H2 H3 + H1 H’2 H3)  =  (

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