## Question 6:

### Part (a):
| $G_v(t) = \dfrac{9}{25}t^2 + \dfrac{12}{25}t^3 + \dfrac{4}{25}t^4$ or $t^2\!\left(\dfrac{9}{25}+\dfrac{12}{25}t+\dfrac{4}{25}t^2\right)$ | M1 | |
| $= t^2\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^2 *$ | A1* cso | |

### Part (b)(i):
| $G_W'(t) = 2t\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^4 + \left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^5$ | M1 | |
| $[G_W'(1) =]\ \mathbf{3}$ | A1 | |

### Part (b)(ii):
| $G_W''(t) = 2\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^4 + \dfrac{16}{5}t\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^3 + 2\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^4$ | M1 | |
| $G_W''(1) = \dfrac{36}{5}$ | A1 | |
| $\text{Var}(W) = "\dfrac{36}{5}" + "3" - ("3")^2$ | M1 | |
| $= \dfrac{6}{5}$ | A1 | |

### Part (c):
| $G_X(t) = t^2\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^2 \times t\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^5$ | M1 | |
| $= t^3\!\left(\dfrac{2}{5}t+\dfrac{3}{5}\right)^7$ | A1 | |

### Part (d):
| $G_Y(t) = t^3 \times (t^2)^3 \times \left(\dfrac{2}{5}t^2+\dfrac{3}{5}\right)^7$ | M1 | |
| $= t^9\!\left(\dfrac{2}{5}t^2+\dfrac{3}{5}\right)^7$ | A1 | |

### Part (e):
| $P(Y=15)$ is coefficient of $t^{15}$ i.e. $\ldots + t^9 \times {}^7C_3\!\left(\dfrac{2}{5}t^2\right)^3\!\left(\dfrac{3}{5}\right)^4+\ldots$ or $P(X=6)$: need coefficient of $t^6$ i.e. $\ldots+t^3\times{}^7C_3\!\left(\dfrac{2}{5}t\right)^3\!\left(\dfrac{3}{5}\right)^4+\ldots$ | M1 | |
| $[P(Y=15) =] \dfrac{22680}{78125} = \dfrac{4536}{15625} = 0.290304$ | A1 | |

# Question (pgf/probability generating functions):

## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct un-simplified pgf based on $\sum t^v P(V=v)$ | M1 | |
| Correct simplified answer | A1* | cso - must see un-simplified version; M1 scored and no incorrect working seen |

## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate using product rule to find $G_W'(t)$, e.g. $5 \times \frac{2}{5}t$ | M1 | Need two terms added, at least one correct; if expanded need 3 correct |
| $3$ from correct derivative | A1 | |

**Alternative:** $W = P+1$ where $P \sim B(5, 0.4)$, so $\text{Var}(W) = \text{Var}(P)$
| $G_P'(t) = 2\left(\frac{2}{5}t + \frac{3}{5}\right)^4$ | M1 | 2.1 |
| $G_W'(1) = 2+1 = 3$ | A1 | 1.1b |

## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt $G_W''(t)$ from their $G_W'(t)$; must be at least 2 terms or a product, one correct term | 1stM1 | Same rule for differentiating a product |
| $\frac{36}{5}$ or $7.2$ from correct derivative | 1stA1 | |
| $G_W''(1) + G_W'(1) - \left(G_W'(1)\right)^2$ from their $G_W''(t)$ if different from $G_W'(t)$ and $G_W(t)$ | 2ndM1 | |
| $\frac{6}{5}$ or $1.2$ | 2ndA1 | Dep on M3A2 |

**Alternative:**
| $G_P''(t) = \frac{16}{5}\left(\frac{2}{5}t+\frac{3}{5}\right)^3$; $G_P''(1) = \frac{16}{5}$ | M1; A1 | 2.1, 1.1b |
| $\text{Var}(W) = \text{"}\frac{16}{5}\text{"}+\text{"2"}-(\text{"2"})^2 = \frac{6}{5}$ | M1; A1 | 2.1, 1.1b |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Realise need to use $G_X(t) = G_V(t) \times G_W(t)$ | M1 | |
| $t^3\left(\frac{2}{5}t+\frac{3}{5}\right)^7$ | A1 | |

## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Realise need to multiply through by $t^3$ or substitute $t^2$ for $t$ in $t^3 G_X(t^2)$ | M1 | |
| $t^9\left(\frac{2}{5}t^2+\frac{3}{5}\right)^7$ | A1 | Need not be in simplest form |

## Part (e):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to find correct coefficient of $t^n$ or identify $Y = 2J+9$ where $J \sim B(7, 0.4)$; expression of form $t^n(at^m+b)^k$ | M1 | Allow statement $P(Y=15)=0$ if it follows from their pgf |
| Correct exact answer or awrt $0.2903$; allow $0.29$ from correct expression | A1 | |

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