A2 June 2024 Q7
7. The probability of winning a prize when playing a single game of Pento is \(\dfrac{1}{5}\)
When more than one game is played the games are independent.
Sam plays 20 games.
Tessa plays a series of games.
Rama invites Sam and Tessa to play some new games of Pento.
They must pay Rama £1 for each game they play but Rama will pay them £2 for the first time they win a prize, £4 for the second time and £\((2w)\) when they win their \(w\)th prize \((w \gt 2)\)
Sam decides to play \(n\) games of Pento with Rama.
Given that Sam chose \(n = 15\)
Tessa agrees to play Pento with Rama. She will play games until she wins \(r\) prizes and then she will stop.
| Scheme | Marks | AO |
|---|---|---|
| [\(X\) = no. of prizes Sam wins] \(\quad X \sim \mathrm{B}(20, 0.2)\) | M1 | 3.3 |
| \(\mathrm{P}(X \geqslant 4) = 1 - \mathrm{P}(X \leqslant 3) = 0.58855\ldots\) awrt 0.589 | A1 | 1.1b |
| (2) |
Notes
M1 for selecting a suitable model (\(\mathrm{B}(20, 0.2)\))
A1 for awrt 0.589
| Scheme | Marks | AO |
|---|---|---|
| [\(Y\) = no. of game when Tessa wins her 4th prize] \(\quad Y \sim \mathrm{negB}(4, 0.2)\) | M1, | 3.3 |
| \(\mathrm{P}(Y = 20) = \dbinom{19}{3}0.2^3 0.8^{16} \times 0.2,\ = 0.043639\ldots\) awrt 0.0436 | A1 | 1.1b |
| (2) |
Notes
M1 for stating correct negative binomial or \({}^{19}C_3\, p^3(1 - p)^{16} \times p\) for some \(p\)
A1 for awrt 0.0436
| Scheme | Marks | AO |
|---|---|---|
| \(S\) = no of prizes Sam wins in \(n\) games \(\qquad S \sim \mathrm{B}(n, 0.2)\) | M1 | 3.1b |
| Profit \(= (2 + 4 + 6 + \ldots + 2S) - n = \dfrac{S}{2}[4 + 2(S - 1)] - n\) | M1 | 2.1 |
| \(= S^2 + S - n\) | A1 | 1.1b |
| \(\mathrm{E}(S) = 0.2n\) and \(\mathrm{Var}(S) = 0.2 \times 0.8n\) | M1 | 3.1b |
| \(\mathrm{E}(S^2) = 0.16n + 0.04n^2\) or \(\dfrac{1}{25}(n^2 + 4n)\) | A1 | 1.1b |
| (*) So expected profit for Sam is \(\dfrac{1}{25}(n^2 + 4n) + \dfrac{1}{5}n - n = \dfrac{1}{25}\left(n^2 - 16n\right)\) | A1cso | 3.2a |
| (6) |
Notes
1st M1 Correct start to problem - sight or use of \(\mathrm{B}(n, 0.2)\). May be implied by \(\mathrm{E}(S) = 0.2n\)
2nd M1 use of AP formula with \(a = d = 2\), or 2×AP formula with \(a = d = 1\), or equivalent
NB: must be working in another variable, AP formulae cannot be in terms of \(n\)
1st A1 \(S^2 + S - n\) or equivalent, must be in a form from which expectation can be found
3rd M1 for use of \(\mathrm{E}(S) = 0.2n\) and \(\mathrm{Var}(S) = 0.16n\) (must be labelled or used as variance) (corrected from the printed mark scheme, which has \(\mathrm{Var}(S^2) = 0.16n\))
2nd A1 for correct value for \(\mathrm{E}(S^2)\)
(*) 3rd A1 for a correct solution only, pulling together everything to get given answer
| Scheme | Marks | AO |
|---|---|---|
| Using profit expression: Require \(\mathrm{P}(S^2 + S - n \geqslant 0)\) Using a listing method: Indicates that 4 wins is first non-loss | M1 | 3.1b |
| Profit expression: Solving quadratic, leading to \(S = \ldots\) Listing: \(\mathrm{P}(S \geqslant 4)\) | M1 | 2.1 |
| \(\mathrm{P}(S \geqslant 4)\) where \(S \sim \mathrm{B}(15, 0.2)\) | M1 | 1.1b |
| \(= 0.35183\ldots\) awrt 0.352 | A1 | 1.1b |
| (4) |
Notes
1st M1 for using a suitable prob statement or using a listing method to indicate first non-loss \(S\)
2nd M1 for solving the inequality or using a listing method to reach \(\mathrm{P}(S \geqslant 4)\)
NB: award first two M marks for \(\mathrm{P}(S \geqslant 4)\) provided it does not come from incorrect working
3rd M1 for attempting \(\mathrm{P}(S \geqslant 4)\) with \(\mathrm{B}(15, 0.2)\)
A1 for awrt 0.352
NB: solutions stemming from finding values of \(n\) gain no marks
| Scheme | Marks | AO |
|---|---|---|
| \(T\) = game on which Tessa wins her \(r\)th prize \(\quad T \sim \mathrm{negB}(r, 0.2)\) or \(r(r + 1)\) | M1 | 3.3 |
| Profit \(= (2 + 4 + 6 + \ldots + 2r) - T = r(r + 1) - T\) | A1 | 1.1b |
| Tessa’s expected profit \(= r(r + 1) - \mathrm{E}(T)\) | M1 | 3.4 |
| \(= r^2 + r - \dfrac{r}{0.2} = \underline{\boldsymbol{r^2 - 4r}}\) | A1 | 1.1b |
| (4) | ||
| (18 marks) |
Notes
1st M1 for sight or use of \(\mathrm{negB}(r, 0.2)\) or sight of \(r(r + 1)\)
1st A1 for \(r(r + 1) - T\), where \(T\) is defined
2nd M1 for use of \(\mathrm{E}(T) = 5r\) in an expression of “revenue” – \(5r\)
2nd A1 for \(r^2 - 4r\)