Let A and B be events such that \({\text{P}}(A) = 0.6,{\text{ P}}(A \cup B) = 0.8{\text{ and P}}(A|B) = 0.6\) .

Find P(B) .

EITHER

Using \({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\)     (M1)

\(0.6{\text{P}}(B) = {\text{P}}(A \cap B)\)     A1

Using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)\) to obtain \(0.8 = 0.6 + {\text{P}}(B) - {\text{P}}(A \cap B)\)     A1

Substituting \(0.6{\text{P}}(B) = {\text{P}}(A \cap B)\) into above equation     M1

OR

As \({\text{P}}(A|B) = {\text{P}}(A)\) then A and B are independent events     M1R1

Using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A) \times {\text{P}}(B)\)     A1

to obtain \(0.8 = 0.6 + {\text{P}}(B) - 0.6 \times {\text{P}}(B)\)     A1

THEN

\(0.8 = 0.6 + 0.4{\text{P}}(B)\)     A1

\({\text{P}}(B) = 0.5\)     A1     N1

[6 marks]

This question was generally well done, with a few candidates spotting an opportunity to use results for the independent events A and B.