Further Mathematics Paper 2, Nov/Dec. 2012
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 7

Eight coins were tossed together several times and the number of times the head appeared was recorded as follows:

 Number of heads 0 1 2 3 4 5 6 7 8 Frequency 3 8 24 37 10 60 79 11 9

Find the probability of obtaining

2. between 2 and 5 heads.

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Observation

The report stated that this question was attempted by majority of the candidates and they performed very well in it.  Candidates were able to obtain the total number of outcomes as   The probability of exactly one head = .
In part (b), candidates should note that between 2 and 5 meant that 2 and 5 were excluded.  Some candidates added the frequency for 2 and 5 and this gave them the wrong answer.  Therefore, required probability = Pr(3) + Pr(4) =  = .
In part (c), probability of at most 1 head = Pr(0) + Pr(1) = .