| Date | November 2017 | Marks available | 1 | Reference code | 17N.2.sl.TZ0.4 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Write down | Question number | 4 | Adapted from | N/A |
Question
A discrete random variable \(X\) has the following probability distribution.
Find the value of \(k\).
Write down \({\text{P}}(X = 2)\).
Find \({\text{P}}(X = 2|X > 0)\).
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)total probability = 1
correct equation (A1)
eg\(\,\,\,\,\,\)\(0.475 + 2{k^2} + \frac{k}{{10}} + 6{k^2} = 1,{\text{ }}8{k^2} + 0.1k - 0.525 = 0\)
\(k = 0.25\) A2 N3
[4 marks]
\({\text{P}}(X = 2) = 0.025\) A1 N1
[1 mark]
valid approach for finding \({\text{P}}(X > 0)\) (M1)
eg\(\,\,\,\,\,\)\(1 - 0.475,{\text{ }}2({0.25^2}) + 0.025 + 6({0.25^2}),{\text{ }}1 - {\text{P}}(X = 0),{\text{ }}2{k^2} + \frac{k}{{10}} + 6{k^2}\)
correct substitution into formula for conditional probability (A1)
eg\(\,\,\,\,\,\)\(\frac{{0.025}}{{1 - 0.475}},{\text{ }}\frac{{0.025}}{{0.525}}\)
0.0476190
\({\text{P}}(X = 2|X > 0) = \frac{1}{{21}}\) (exact), 0.0476 A1 N2
[3 marks]
