(a) | $[X = \text{no. of prizes Sam wins}] \quad X \sim B(20, 0.2)$ | M1 | 3.3 |
| $P(X \geq 4) = 1 - P(X \leq 3) = 0.58855\ldots$ awrt $\mathbf{0.589}$ | A1 | 1.1b |

(b) | $[Y = \text{no. of game when Tessa wins her 4}^{\text{th}} \text{ prize}] \quad Y \sim \text{negB}(4, 0.2)$ | M1 | 3.3 |
| $P(Y = 20) = \binom{19}{3}0.2^3 \times 0.8^{16} \times 0.2, = 0.043639\ldots$ awrt $\mathbf{0.0436}$ | A1 | 1.1b |

(c) | $S = \text{no of prizes Sam wins in } n \text{ games}$ $\quad S \sim B(n, 0.2)$ | M1 | 3.1b |
| Profit $= (2 + 4 + 6 + \ldots + 2S) - n = \sum_{i=1}^{2(S-1)} [-n] = S^2 + S - n$ | M1 | 2.1 |
| $E(S) = 0.2n$ and $\text{Var}(S) = 0.2 \times 0.8n$ | A1, M1 | 1.1b; 3.1b |
| $E(S^2) = 0.16n + 0.04n^2 + \frac{1}{25}(n^2 + 4n)$ | A1 | 1.1b |
| $(*) \text{ So expected profit for Sam is } \frac{1}{25}(n^2 + 4n) + \frac{1}{5}n - n = \frac{1}{25}(n^2 - 16n)$ | A1cso | 3.2a |

(d) | **Using profit expression:** | M1 | 3.1b |
| Require $P(S^2 + S - n \geq 0)$ |  |  |
| Solving quadratic, leading to $S \geq \ldots$ | M1 | 2.1 |
| $P(S \geq 4)$ where $S \sim B(15, 0.2)$ | M1 | 1.1b |
| $= 0.35183\ldots$ awrt $\mathbf{0.352}$ | A1 | 1.1b |
| **Using a listing method** | M1 | 3.1b |
| Indicates that 4 wins is first non-loss | M1 | 2.1 |
| $P(S \geq 4)$ where $S \sim B(15, 0.2)$ | M1 | 1.1b |
| $= 0.35183\ldots$ awrt $\mathbf{0.352}$ | A1 | 1.1b |

(e) | $T = \text{game on which Tessa wins her } r^{\text{th}} \text{ prize}$ $T \sim \text{negB}(r, 0.2)$ 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 $= r(r+1) - E(T) = r^2 + r - \frac{r}{0.2} = r^2 - 4r$ | M1 | 3.4 |
| $= r^2 - 4r$ | A1 | 1.1b |

**Notes:**
- (a) M1 for selecting a suitable model (B(20, 0.2))
- (a) A1 for awrt 0.589
- (b) M1 for stating correct negative binomial or 19C3p³(1−p)¹⁶×p for some p
- (b) A1 for awrt 0.0436
- (c) 1st M1 Correct start to problem - sight or use of B(n, 0.2). May be implied by E(S) = 0.2n
- (c) 2nd M1 use of AP formula with $a = d = 2$, or 2xAP formula with $a = d = 1$, or equivalent
- (c) **NB:** must be working in another variable, AP formulae cannot be in terms of n
- (c) 1st A1 for $S^2 + S - n$ or equivalent, must be in a form from which expectation can be found
- (c) 3rd M1 for use of $E(S) = 0.2n$ and Var$(S^2) = 0.16n$ (must be labelled or used as variance)
- (c) 2nd A1 for correct unsimplified $E(S^2)$ – values to 2 s.f. or better
- (c) 3rd A1 for a correct solution only, pulling together everything to get given answer
- (*) **3rd A1** for a correct solution only, pulling together everything to get given answer
- (d) 1st M1 for using a suitable prob statement or using a listing method to indicate first non-loss S
- (d) 2nd M1 for solving the inequality or using a listing method to reach $P(S \geq 4)$
- **(d) NB:** award first two M marks for $P(S \geq 4)$ provided it does not come from incorrect working
- (d) 3rd M1 for attempting $P(S \geq 4)$ with B(15, 0.2)
- (d) A1 for awrt 0.352
- **(d) NB:** solutions stemming from finding values of n gain no marks
- (e) 1st M1 for sight or use of negB(r, 0.2) or sight of $r(r+1)$
- (e) 1st A1 for $r(r+1) - T$, where $T$ is defined
- (e) 2nd M1 for use of $E(T) = 5r$ in an expression of "revenue" − 5r
- (e) 2nd A1 for $r^2 - 4r$

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