| Date | None Specimen | Marks available | 6 | Reference code | SPNone.3sp.hl.TZ0.2 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Show that | Question number | 2 | Adapted from | N/A |
Question
When Andrew throws a dart at a target, the probability that he hits it is \(\frac{1}{3}\) ; when Bill throws a dart at the target, the probability that he hits the it is \(\frac{1}{4}\) . Successive throws are independent. One evening, they throw darts at the target alternately, starting with Andrew, and stopping as soon as one of their darts hits the target. Let X denote the total number of darts thrown.
Write down the value of \({\text{P}}(X = 1)\) and show that \({\text{P}}(X = 2) = \frac{1}{6}\).
Show that the probability generating function for X is given by
\[G(t) = \frac{{2t + {t^2}}}{{6 - 3{t^2}}}.\]
Hence determine \({\text{E}}(X)\).
Markscheme
\({\text{P}}(X = 1) = \frac{1}{3}\) A1
\({\text{P}}(X = 2) = \frac{2}{3} \times \frac{1}{4}\) A1
\(= \frac{1}{6}\) AG
[2 marks]
\(G(t) = \frac{1}{3}t + \frac{2}{3} \times \frac{1}{4}{t^2} + \frac{2}{3} \times \frac{3}{4} \times \frac{1}{3}{t^3} + \frac{2}{3} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{4}{t^4} + \ldots \) M1A1
\( = \frac{1}{3}t\left( {1 + \frac{1}{2}{t^2} + \ldots } \right) + \frac{1}{6}{t^2}\left( {1 + \frac{1}{2}{t^2} + \ldots } \right)\) M1A1
\( = \frac{{\frac{t}{3}}}{{1 - \frac{{{t^2}}}{2}}} + \frac{{\frac{{{t^2}}}{6}}}{{1 - \frac{{{t^2}}}{2}}}\) A1A1
\( = \frac{{2t + {t^2}}}{{6 - 3{t^2}}}\) AG
[6 marks]
\(G'(t) = \frac{{(2 + 2t)(6 - 3{t^2}) + 6t(2t + {t^2})}}{{{{(6 - 3{t^2})}^2}}}\) M1A1
\({\text{E}}(X) = G'(1) = \frac{{10}}{3}\) M1A1
[4 marks]
