| Date | November 2016 | Marks available | 9 | Reference code | 16N.3sp.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Calculate | Question number | 3 | Adapted from | N/A |
Question
Alun answers mathematics questions and checks his answer after doing each one.
The probability that he answers any question correctly is always \(\frac{6}{7}\), independently of all other questions. He will stop for coffee immediately following a second incorrect answer. Let \(X\) be the number of questions Alun answers before he stops for coffee.
Nic answers mathematics questions and checks his answer after doing each one.
The probability that he answers any question correctly is initially \(\frac{6}{7}\). After his first incorrect answer, Nic loses confidence in his own ability and from this point onwards, the probability that he answers any question correctly is now only \(\frac{4}{7}\).
Both before and after his first incorrect answer, the result of each question is independent of the result of any other question. Nic will also stop for coffee immediately following a second incorrect answer. Let \(Y\) be the number of questions Nic answers before he stops for coffee.
(i) State the distribution of \(X\), including its parameters.
(ii) Calculate \({\text{E}}(X)\).
(iii) Calculate \({\text{P}}(X = 5)\).
(i) Calculate \({\text{E}}(Y)\).
(ii) Calculate \({\text{P}}(Y = 5)\).
Markscheme
(i) \({\text{NB}}\left( {2,{\text{ }}\frac{1}{7}} \right)\) A1A1A1
Note: The final A1 mark can be awarded for knowing that \(p = \frac{1}{7}\) independent of the other two marks.
(ii) \({\text{E}}(X) = \frac{r}{p} = 14\) A1
(iii) \(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right){\left( {\frac{6}{7}} \right)^3}{\left( {\frac{1}{7}} \right)^2} = 0.0514\) (M1)A1
Note: Accept any number that rounds to this 3sf number.
[6 marks]
(i) \(Y = {Y_1} + {Y_2}\) (number up to1st + number up to 2nd) (M1)
\({Y_1} \sim Geo\left( {\frac{1}{7}} \right),{\text{ }}{Y_2} \sim Geo\left( {\frac{3}{7}} \right)\) (A1)
Notes: The above (A1) is independent of the (M1).
Could have \({\text{NB }}(1,{\text{ }}p)\), instead of \(Geo(p)\).
\({\text{E}}(Y) = \frac{1}{{\left( {\frac{1}{7}} \right)}} + \frac{1}{{\left( {\frac{3}{7}} \right)}} = 7 + \frac{7}{3} = 9\frac{1}{3}{\text{ (9.33)}}\) M1A1
(ii) \(Y = {Y_1} + {Y_2} = 5\) happens when (M1)
\({Y_1} = 1,{\text{ }}{Y_2} = 4\) or \({Y_1} = 2,{\text{ }}{Y_2} = 3\) or \({Y_1} = 3,{\text{ }}{Y_2} = 2\) or \({Y_1} = 4,{\text{ }}{Y_2} = 1\) (A1)
so probability is \(\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{3}{7}\) (M1)(A1)
\( = 0.0928{\text{ }}\left( {\frac{{1560}}{{16807}}} \right)\) A1
Note: Accept any answer that rounds to 0.093.
[9 marks]
