Show that the probability generating function of X is given by

\[G\left( t \right) = \frac{{pt}}{{1 - qt}}.\]

Deduce that \({\text{E}}\left( X \right) = \frac{1}{p}\).

Two friends A and B play a ball game with the following rules.

Each player starts with zero points. Player A serves first and then the players have alternate pairs of serves so that the service order is A, B, B, A, A, … When player A serves, the probability of her scoring 1 point is \({p_A}\) and the probability of B scoring 1 point is \({q_A}\), where \({q_A} = 1 - {p_A}\).

When player B serves, the probability of her scoring 1 point is \({p_B}\) and the probability of A scoring 1 point is \({q_B}\), where \({q_B} = 1 - {p_B}\).

Show that, after the first 6 serves, the probability that each player has 3 points is

\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}
3 \hfill \\
x \hfill \\
\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 - x}}{\left( {{q_B}} \right)^{3 - x}}\).

After 6 serves the score is 3 points each. Play continues and the game ends when one player has scored two more points than the other player. Let N be the number of further serves required before the game ends. Given that \({p_A}\) = 0.7 and \({p_A}\) = 0.6 find P(N = 2).

Let \(M = \frac{1}{2}N\). Show that M has a geometric distribution and hence find the value of E(N).

\({\text{P}}\left( {X = x} \right) = p{q^{x - 1}},\,{\text{for}}\,x = 1,\,2 \ldots \)

\(G\left( t \right) = \sum\limits_{x = 1}^\infty  {{t^x}} p{q^{x - 1}}\)      M1

\( = pt\sum\limits_{x = 1}^\infty  {{{\left( {tq} \right)}^{x - 1}}} \)       A1

\( = pt\left( {1 + tq + {{\left( {tq} \right)}^2} \ldots } \right)\)      M1

\( = \frac{{pt}}{{1 - tq}}\)      AG

[3 marks]

\(G'\left( t \right) = \frac{{\left( {1 - tq} \right)p - pt\left( { - q} \right)}}{{{{\left( {1 - tq} \right)}^2}}}\)      M1A1

\({\text{E}}\left( X \right) = G'\left( 1 \right)\)      M1

\( = \frac{{\left( {1 - q} \right)p + pq}}{{{{\left( {1 - q} \right)}^2}}}\)      A1

\( = \frac{1}{p}\)        AG

[4 marks]

after 6 serves (3 serves each) we have ABBAAB

A serves   B serves

3 wins       0 losses      \({p_1} = {}^3{C_3}p_A^3q_A^0{}^3{C_0}p_B^3q_B^0\)     M1A1

2 wins       1 loss         \({p_2} = {}^3{C_2}p_A^2q_A^1{}^3{C_1}p_B^2q_B^1\)      A1

1 win         2 losses     \({p_3} = {}^3{C_1}p_A^1q_A^2{}^3{C_2}p_B^1q_B^2\)      A1

0 wins       3 losses     \({p_4} = {}^3{C_0}p_A^0q_A^3{}^3{C_3}p_B^0q_B^3\)       A1

since \({}^3{C_0} = {}^3{C_3},\,\,{}^3{C_1} = {}^3{C_2}\)

\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}
3 \hfill \\
x \hfill \\
\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 - x}}{\left( {{q_B}} \right)^{3 - x}}\)     AG

[5 marks]

for N = 2 serves are B, A respectively

P(N = 2) = P(B wins twice) + P(A wins twice)      (M1)

= 0.6 × 0.3 + 0.4 × 0.7       A1

= 0.46       A1

[3 marks]

for \(M = \frac{1}{2}N\)

\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M}\)      M1

\({\text{P}}\left( {M = 2} \right) = {\text{P}}\left( {N = 4} \right)\)

\( = {\text{P}}\left( {\begin{array}{*{20}{c}}
{{\text{game does not end after}}} \\
{{\text{first two serves}}}
\end{array}} \right) \times {\text{P}}\left( {\begin{array}{*{20}{c}}
{{\text{game ends after}}} \\
{{\text{next two serves}}}
\end{array}} \right) = \left( {1 - {p_M}} \right){p_M}\)      A1

similarly \({\text{P}}\left( {M = 3} \right) = {\left( {1 - {p_M}} \right)^2}{p_M}\)      (A1)

hence \({\text{P}}\left( {M = r} \right) = {\left( {1 - {p_M}} \right)^{r - 1}}{p_M}\)       A1

hence M has a geometric distribution      AG

\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M} = 0.46\)      A1

hence \({\text{E}}\left( M \right) = \frac{1}{p} = \frac{1}{{0.46}} = 2.174\)

\({\text{E}}\left( N \right) = {\text{E}}\left( {2M} \right) = 2{\text{E}}\left( M \right)\)       M1

= 4.35      A1

[7 marks]