| Date | May 2008 | Marks available | 10 | Reference code | 08M.3sp.hl.TZ1.5 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ1 |
| Command term | Find, Show that, State, and Hence | Question number | 5 | Adapted from | N/A |
Question
Anna has a fair cubical die with the numbers 1, 2, 3, 4, 5, 6 respectively on the six faces. When she tosses it, the score is defined as the number on the uppermost face. One day, she decides to toss the die repeatedly until all the possible scores have occurred at least once.
(a) Having thrown the die once, she lets \({X_2}\) denote the number of additional throws required to obtain a different number from the one obtained on the first throw. State the distribution of \({X_2}\) and hence find \({\text{E}}({X_2})\) .
(b) She then lets \({X_3}\) denote the number of additional throws required to obtain a different number from the two numbers already obtained. State the distribution of \({X_3}\) and hence find \({\text{E}}({X_3})\) .
(c) By continuing the process, show that the expected number of tosses needed to obtain all six possible scores is 14.7.
Markscheme
(a) \({X_2}\) is a geometric random variable A1
with \(p = \frac{5}{6}.\) A1
Therefore \({\text{E}}({X_2}) = \frac{6}{5}.\) A1
[3 marks]
(b) \({X_3}\) is a geometric random variable with \(p = \frac{4}{6}.\) A1
Therefore \({\text{E}}({X_3}) = \frac{6}{4}.\) A1
[2 marks]
(c) \({\text{E}}({X_4}) = \frac{6}{3},{\text{ E}}({X_5}) = \frac{6}{2},{\text{ E}}({X_6}) = \frac{6}{1}\) A1A1A1
\({\text{E}}({X_1}) = 1\,\,\,\,\,{\text{(or }}{X_1} = 1)\) A1
Expected number of tosses \(\sum\limits_{n = 1}^6 {{\text{E}}({X_n})} \) M1
\( = 14.7\) AG
[5 marks]
Total [10 marks]
Examiners report
Many candidates were unable even to start this question although those who did often made substantial progress.
