| Date | May 2016 | Marks available | 2 | Reference code | 16M.1.hl.TZ2.7 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Show that | Question number | 7 | Adapted from | N/A |
Question
\(A\) and \(B\) are independent events such that \({\text{P}}(A) = {\text{P}}(B) = p,{\text{ }}p \ne 0\).
Show that \({\text{P}}(A \cup B) = 2p - {p^2}\).
Find \({\text{P}}(A|A \cup B)\) in simplest form.
Markscheme
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)\)
\( = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A){\text{P}}(B)\) (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]
Examiners report
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.
