Date | November 2015 | Marks available | 2 | Reference code | 15N.2.sl.TZ0.5 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Write down | Question number | 5 | Adapted from | N/A |
Question
Let \(C\) and \(D\) be independent events, with \({\text{P}}(C) = 2k\) and \({\text{P}}(D) = 3{k^2}\), where \(0 < k < 0.5\).
Write down an expression for \({\text{P}}(C \cap D)\) in terms of \(k\).
Given that \({\text{P}}(C \cap D) = 0.162\) find \(k\).
Find \({\text{P}}(C'|D)\).
Markscheme
\({\text{P}}(C \cap D) = 2k \times 3{k^2}\) (A1)
\({\text{P}}(C \cap D) = 6{k^3}\) A1 N2
[2 marks]
their correct equation (A1)
eg\(\;\;\;2k \times 3{k^2} = 0.162,{\text{ }}6{k^3} = 0.162\)
\(k = 0.3\) A1 N2
METHOD 1
finding their \({\text{P}}(C' \cap D)\) (seen anywhere) (A1)
eg \(0.4 \times 0.27,0.27 - 0.162,0.108\)
correct substitution into conditional probability formula (A1)
eg\(\;\;\;{\text{P}}(C'|D) = \frac{{{\text{P}}(C' \cap D)}}{{0.27}},{\text{ }}\frac{{(1 - 2k)(3{k^2})}}{{3{k^2}}}\)
\({\text{P}}(C'|D) = 0.4\) A1 N2
METHOD 2
recognizing \({\text{P}}(C'|D) = {\text{P}}(C')\) A1
finding their \({\text{P}}(C') = 1 - {\text{P}}(C)\) (only if first line seen) (A1)
eg\(\;\;\;1 - 2k,{\text{ }}1 - 0.6\)
\({\text{P}}(C'|D) = 0.4\) A1 N2
[3 marks]
Total [7 marks]