| Date | May 2011 | Marks available | 4 | Reference code | 11M.1.hl.TZ1.1 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.3\) and \({\text{P}}(B) = 0.4\) .
Find the value of \({\text{P}}(A \cup B)\) when
(i) \(A\) and \(B\) are mutually exclusive;
(ii) \(A\) and \(B\) are independent.
Given that \({\text{P}}(A \cup B) = 0.6\) , find \({\text{P}}(A|B)\) .
Markscheme
(i) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) = 0.7\) A1
(ii) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)\) (M1)
\( = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A){\text{P}}(B)\) (M1)
\( = 0.3 + 0.4 - 0.12 = 0.58\) A1
[4 marks]
\({\text{P}}(A \cap B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cup B)\)
\( = 0.3 + 0.4 - 0.6 = 0.1\) A1
\({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\) (M1)
\( = \frac{{0.1}}{{0.4}} = 0.25\) A1
[3 marks]
Examiners report
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
