| Date | November 2017 | Marks available | 2 | Reference code | 17N.2.hl.TZ0.2 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Show that and Hence | Question number | 2 | Adapted from | N/A |
Question
Events \(A\) and \(B\) are such that \({\text{P}}(A \cup B) = 0.95,{\text{ P}}(A \cap B) = 0.6\) and \({\text{P}}(A|B) = 0.75\).
Find \({\text{P}}(B)\).
Find \({\text{P}}(A)\).
Hence show that events \(A’\) and \(B\) are independent.
Markscheme
\({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\)
\( \Rightarrow 0.75 = \frac{{0.6}}{{{\text{P}}(B)}}\) (M1)
\( \Rightarrow {\text{P}}(B){\text{ }}\left( { = \frac{{0.6}}{{0.75}}} \right) = 0.8\) A1
[2 marks]
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)\)
\( \Rightarrow 0.95 = {\text{P}}(A) + 0.8 - 0.6\) (M1)
\( \Rightarrow {\text{P}}(A) = 0.75\) A1
[2 marks]
METHOD 1
\({\text{P}}(A'|B) = \frac{{{\text{P}}(A' \cap B)}}{{{\text{P}}(B)}} = \frac{{0.2}}{{0.8}} = 0.25\) A1
\({\text{P}}(A'|B) = {\text{P}}(A’)\) R1
hence \(A’\) and \(B\) are independent AG
Note: If there is evidence that the student has calculated \({\text{P}}(A' \cap B) = 0.2\) by assuming independence in the first place, award A0R0.
METHOD 2
EITHER
\({\text{P}}(A) = {\text{P}}(A|B)\) A1
OR
\({\text{P}}(A) \times {\text{P}}(B) = 0.75 \times 0.80 = 0.6 = {\text{P}}(A \cap B)\) A1
THEN
\(A\) and \(B\) are independent R1
hence \(A’\) and \(B\) are independent AG
METHOD 3
\({\text{P}}(A') \times {\text{P}}(B) = 0.25 \times 0.80 = 0.2\) A1
\({\text{P}}(A') \times {\text{P}}(B) = {\text{P}}(A' \cap B)\) R1
hence \(A’\) and \(B\) are independent AG
[2 marks]
