| Date | May 2015 | Marks available | 4 | Reference code | 15M.2.hl.TZ0.6 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Show that | Question number | 6 | Adapted from | N/A |
Question
Gillian is throwing a ball at a target. The number of throws she makes before hitting the target follows a geometric distribution, \(X \sim {\text{Geo}}(p)\). When she uses a cricket ball the number of throws she makes follows a geometric distribution with \(p = \frac{1}{4}\). When she uses a tennis ball the number of throws she makes follows a geometric distribution with \(p = \frac{3}{4}\). There is a box containing a large number of balls, \(80\%\) of which are cricket balls and the remainder are tennis balls. The random variable \(A\) is the number of throws needed to hit the target when a single ball is chosen at random from this box and used for all throws.
Find \({\text{E}}(A)\).
Show that \({\text{P}}(A = r) = \frac{1}{5} \times {\left( {\frac{3}{4}} \right)^{r - 1}} + \frac{3}{{20}} \times {\left( {\frac{1}{4}} \right)^{r - 1}}\).
Find \({\text{P}}(A \leqslant 5|A > 3)\).
Markscheme
\({\text{E}}(X{\text{ }}tennis) = \frac{1}{{\frac{3}{4}}} = \frac{4}{3}\) (A1)
\({\text{E}}(X{\text{ }}cricket) = \frac{1}{{\frac{1}{4}}} = 4\) (A1)
\({\text{E}}(A) = \frac{4}{3} \times \frac{1}{5} + 4 \times \frac{4}{5} = \frac{{52}}{{15}}\) M1A1
\({\text{P}}(X = r|cricket) = \frac{1}{4} \cdot {\left( {\frac{3}{4}} \right)^{r - 1}}\) (A1)
\({\text{P}}(X = r|tennis) = \frac{3}{4} \cdot {\left( {\frac{1}{4}} \right)^{r - 1}}\) (A1)
\({\text{P}}(A = r) = {\text{P}}(X = r|cricket) \times {\text{P}}(cricket) + {\text{P}}(X = r|tennis) \times {\text{P}}(tennis)\) (M1)
\( = \frac{1}{4} \times {\left( {\frac{3}{4}} \right)^{r - 1}} \times \frac{4}{5} + \frac{3}{4} \times {\left( {\frac{1}{4}} \right)^{r - 1}} \times \frac{1}{5}\) A1
\( = \frac{1}{5} \times {\left( {\frac{3}{4}} \right)^{r - 1}} + \frac{3}{{20}} \times {\left( {\frac{1}{4}} \right)^{r - 1}}\) AG
\({\text{P}}(A \leqslant 5)|(A < 3) = \frac{{{\text{P}}(A = 4{\text{ or }}5)}}{{1 - {\text{P}}(A = 1{\text{ or 2 or }}3)}}\) M1A1A1
\({\text{P}}(A = 1) = \frac{7}{{20}}\)
\({\text{P}}(A = 2) = \frac{15}{{80}}\)
\({\text{P}}(A = 3) = \frac{39}{{320}}\)
\({\text{P}}(A = 4) = \frac{111}{{1280}}\)
\({\text{P}}(A = 5) = \frac{327}{{5120}}\)
\( \Rightarrow {\text{P}}(A > 3)|(A \leqslant 5) = \frac{{\frac{{771}}{{5120}}}}{{\frac{{1744}}{{5120}}}}\) A1A2
Note: Award A1 for correct working for numerator and Award A2 for correct working for denominator
\( = \frac{{771}}{{1744}}\;\;\;( = 0.442)\) A1
Examiners report
It was pleasing to see many correct answers to parts a) and b) with candidates correctly recognising how to work with the distribution.
It was pleasing to see many correct answers to parts a) and b) with candidates correctly recognising how to work with the distribution.
Part c) caused more problems. Although a number of wholly correct solutions were seen, many candidates were unable to work meaningfully with the conditional probability.
