Further Mathematics Paper 2, Nov/Dec. 2011
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 5

X and Y are two events such that P(X  Y) =  and P(X) = . Find P(Y) if the events are

1. mutually exclusive
2. independent
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Observation

The report stated that this was another question where majority of the candidates attempted it and performed creditably. However, it was also observed that candidates’ performance in part (a) was better than it was in part (b). Some of them were able to determine P(Y) when the events were mutually exclusive but could not determine P(Y) when they were independent.
In part (a), when the events were mutually exclusive, P(Y) was obtained as P(X  Y) – P(X)
=  -  = .
In part (b), since the events were independent, then P(X  Y) = P(X) x P(Y). P(Y) was then obtained as  P(Y) = P(X  Y) – P(X) + P(X  Y) i.e. P(Y) =  -  + (P(Y)).  = . Therefore, P(Y) =  x  = .