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.