Two events A and B are such that \({\text{P}}(A \cup B) = 0.7\) and \({\text{P}}(A|B') = 0.6\).

Find \({\text{P}}(B)\).

Note: Be aware that an unjustified assumption of independence will also lead to P(B) = 0.25, but is an invalid method.

 

METHOD 1

\({\text{P}}(A'|B') = 1 - {\text{P}}(A|B') = 1 - 0.6 = 0.4\)     M1A1

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

\({\text{P}}(A' \cap B') = {\text{P}}\left( {(A \cup B)'} \right) = 1 - 0.7 = 0.3\)     A1

\(0.4 = \frac{{0.3}}{{{\text{P}}(B')}} \Rightarrow {\text{P(}}B') = 0.75\)     (M1)A1

\({\text{P}}(B) = 0.25\)     A1

(this method can be illustrated using a tree diagram)

[6 marks]

 

METHOD 2

\({\text{P}}\left( {(A \cup B)'} \right) = 1 - 0.7 = 0.3\)     A1

\({\text{P}}(A|B') = \frac{x}{{x + 0.3}} = 0.6\)     M1A1

\(x = 0.6x + 0.18\)

\(0.4x = 0.18\)

\(x = 0.45\)     A1

\({\text{P}}(A \cup B) = x + y + z\)

\({\text{P}}(B) = y + z = 0.7 - 0.45\)     (M1)

\( = 0.25\)     A1

[6 marks]

 

METHOD 3

\(\frac{{{\text{P}}(A \cap B')}}{{{\text{P}}(B')}} = 0.6{\text{ (or P}}(A \cap B') = 0.6{\text{P}}(B')\)     M1

\({\text{P}}(A \cap B') = {\text{P}}(A \cup B) - {\text{P}}(B)\)     M1A1

\({\text{P}}(B') = 1 - {\text{P}}(B)\)

\(0.7 - {\text{P}}(B) = 0.6 - 0.6{\text{P}}(B)\)     M1(A1)

\(0.1 = 0.4{\text{P}}(B)\)

\({\text{P}}(B) = \frac{1}{4}\)     A1

[6 marks]

There is a great variety of ways to approach this question and there were plenty of very good solutions produced, all of which required an insight into the structure of conditional probability. A few candidates unfortunately assumed independence and so did not score well.