| Date | May 2010 | Marks available | 2 | Reference code | 10M.1.sl.TZ2.8 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Let \({\text{P}}(A) = 0.5\), \({\text{P}}(B) = 0.6\) and \({\text{P}}(A \cup B) = 0.8\).
Find \({\text{P}}(A \cap B)\).
Find \({\text{P}}(A|B)\).
Decide whether A and B are independent events. Give a reason for your answer.
Markscheme
\(0.8 = 0.5 + 0.6 - {\text{P}}(A \cap B)\) (M1)
\({\text{P}}(A \cap B) = 0.3\) (A1) (C2)
Note: Award (M1) for correct substitution, (A1) for correct answer.
[2 marks]
\({\text{P}}(A|B) = \frac{{0.3}}{{0.6}}\) (M1)
= 0.5 (A1)(ft) (C2)
Note: Award (M1) for correct substitution in conditional probability formula. Follow through from their answer to part (a), provided probability is not greater than one.
[2 marks]
\({\text{P}}(A \cap B) = {\text{P}}(A) \times {\text{P}}(B)\) or 0.3 = 0.5 × 0.6 (R1)
OR
\({\text{P}}(A|B) = {\text{P}}(A)\) (R1)
they are independent. (Yes) (A1)(ft) (C2)
Note: Follow through from their answers to parts (a) or (b).
Do not award (R0)(A1).
[2 marks]
Examiners report
Parts (a) and (b) were well answered but very few candidates could provide a reason for the independence of A and B. A number of candidates confused independent and mutually exclusive events.
Parts (a) and (b) were well answered but very few candidates could provide a reason for the independence of A and B. A number of candidates confused independent and mutually exclusive events.
Parts (a) and (b) were well answered but very few candidates could provide a reason for the independence of A and B. A number of candidates confused independent and mutually exclusive events.
