(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 ${\text{P}}(Y = 5)$.

(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]