Show that ${\text{P}}(A \cup B) = 2p - {p^2}$.

Find ${\text{P}}(A|A \cup B)$ in simplest form.

${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)$

$= {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A){\text{P}}(B)$    (M1)

$= p + p - {p^2}$    A1

$= 2p - {p^2}$    AG

[2 marks]

${\text{P}}(A|A \cup B) = \frac{{{\text{P}}\left( {A \cap (A \cup B)} \right)}}{{{\text{P}}(A \cup B)}}$    (M1)

Note:     Allow ${\text{P}}(A \cap A \cup B)$ if seen on the numerator.

$= \frac{{{\text{P}}(A)}}{{{\text{P}}(A \cup B)}}$    (A1)

$= \frac{p}{{2p - {p^2}}}$    A1

$= \frac{1}{{2 - p}}$    A1

[4 marks]

Part (a) posed few problems. Part (b) was possibly a good discriminator for the 4/5 candidates. Some were aware of an alternative (useful) form for the conditional probability, but were unable to interpret $P\left( {A \cap (A \cup B)} \right)$. Large numbers of fully correct answers were seen.

Part (a) posed few problems. Part (b) was possibly a good discriminator for the 4/5 candidates. Some were aware of an alternative (useful) form for the conditional probability, but were unable to interpret $P\left( {A \cap (A \cup B)} \right)$. Large numbers of fully correct answers were seen.