A2 June 2019 Q7
7. A spinner can land on red or blue. When the spinner is spun, there is a probability of \(\dfrac{1}{3}\) that it lands on blue. The spinner is spun repeatedly.
The random variable \(B\) represents the number of the spin when the spinner first lands on blue.
Steve invites Tamara to play a game with this spinner.
Tamara must choose a colour, either red or blue.
Steve will spin the spinner repeatedly until the spinner first lands on the colour Tamara has chosen. The random variable \(X\) represents the number of the spin when this occurs.
If Tamara chooses red, her score is \(\mathrm{e}^X\)
If Tamara chooses blue, her score is \(X^2\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\left[B \sim \mathrm{Geo}\left(\tfrac{1}{3}\right)\right]\) \(\mathrm{P}(B = 4) = \left(\tfrac{2}{3}\right)^3 \times \tfrac{1}{3}\) | M1 | 3.3 |
| \(= \underline{\tfrac{8}{81}}\) | A1 | 1.1b |
| (ii) \(\mathrm{P}(B \leqslant 5) = 1 - \mathrm{P}(B \gt 5)\) or \(1 - \left(\tfrac{2}{3}\right)^5\) | M1 | 2.1 |
| \(= \underline{\tfrac{211}{243}}\) | A1 | 1.1b |
| (4) |
Notes
(i) M1 for selecting the correct model i.e. \(\mathrm{Geo}(p)\) (May be implied by a correct expression)
A1 for \(\frac{8}{81}\) (\(= 0.098765\ldots\) accept awrt 0.0988)
(ii) M1 for a suitable strategy to use the geometric model to find a correct expression
A1 for \(\frac{211}{243}\) (\(= 0.868312\ldots\) accept awrt 0.868)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(B^2) = \mathrm{Var}(B) + \left[\mathrm{E}(B)\right]^2\) | M1 | 2.1 |
| From formula booklet: \(\mathrm{E}(B) = \dfrac{1}{\frac{1}{3}} = 3\) and \(\mathrm{Var}(B) = \dfrac{1 - \frac{1}{3}}{\left(\frac{1}{3}\right)^2} = 6\) | B1 | 1.1b |
| So \(\mathrm{E}(B^2) = 6 + 9 = \underline{\mathbf{15}}\) | A1 | 1.1b |
| (3) |
Notes
M1 for a suitable strategy to find \(\mathrm{E}(B^2)\) [allow \(\mathrm{G}^{\prime\prime}(1) + \mathrm{G}^{\prime}(1)\)]
B1 for use of the correct formulae to find \(\mathrm{E}(B) = 3\) and \(\mathrm{Var}(B) = 6\) or \(\mathrm{G}^{\prime\prime}(1) = 12\)
A1 for 15
SC Formula for \(\mathrm{E}(B^2)\) Allow M1B1A0 for \(\mathrm{E}(B^2) = \dfrac{2 - p}{p^2}\) (o.e.)
| Scheme | Marks | AO |
|---|---|---|
| [Let \(R\) = no. of the spin when it first lands on red] \(X = R \sim \mathrm{Geo}\left(\tfrac{2}{3}\right)\) | M1 | 3.3 |
| Require \(\mathrm{E}(\mathrm{e}^X) = \displaystyle\sum_{x=1}^{\infty} \mathrm{e}^x \left(\tfrac{1}{3}\right)^{x-1} \tfrac{2}{3}\) | M1 | 3.1a |
| \(= \dfrac{2\mathrm{e}}{3} \displaystyle\sum_{x=1}^{\infty} \left(\tfrac{\mathrm{e}}{3}\right)^{x-1}\) | M1 | 2.1 |
| \(= \dfrac{2\mathrm{e}}{3} \times \dfrac{1}{1 - \frac{\mathrm{e}}{3}}\) or \(\dfrac{2\mathrm{e}}{3 - \mathrm{e}}\) | A1 | 1.1b |
| \(\mathrm{E}(\mathrm{e}^X) = 19.297\ldots\) {\(\gt 15 = \mathrm{E}(B^2)\)} so Tamara should choose red since it has the greater expected score | A1 | 2.2a |
| (5) | ||
| (12 marks) |
Notes
1st M1 for choosing a suitable geometric model (sight of \(\mathrm{Geo}\left(\frac{2}{3}\right)\) or at least 3 correct probabilities)
2nd M1 for realising the need for appropriate expected value and using \(\mathrm{E}(\mathrm{g}(X))\) [Need sum and \(\mathrm{f}(x)\)]
NB simply finding \(\mathrm{e}^{\mathrm{E}(X)} = \mathrm{e}^{1.5} =\) awrt 4.48 is M0 and probably no more marks.
3rd M1 for a suitable strategy to turn the expression into a sum that can be found
1st A1 for correct use of sum to infinity of geometric series
2nd A1 for interpreting the outcome of the calculations in terms of a solution to the problem must choose red and see the awrt 19.3 (and allow ft of their \(\mathrm{E}(B^2) \lt 19\))