Determine the probability generating function for $X \sim {\text{B}}(1,{\text{ }}p)$.

Explain why the probability generating function for ${\text{B}}(n,{\text{ }}p)$ is a polynomial of degree $n$.

Two independent random variables ${X_1}$ and ${X_2}$ are such that ${X_1} \sim {\text{B}}(1,{\text{ }}{p_1})$ and ${X_2} \sim {\text{B}}(1,{\text{ }}{p_2})$. Prove that if ${X_1} + {X_2}$ has a binomial distribution then ${p_1} = {p_2}$.

${\text{P}}(X = 0) = 1 - p( = q);{\text{ P}}(X = 1) = p$     (M1)(A1)

${{\text{G}}_x}(t) = \sum\limits_r {{\text{P}}(X = r){t^r}\;\;\;}$(or writing out term by term)     M1

$= q + pt$     A1

[4 marks]

METHOD 1

$PGF$ for $B(n,{\text{ }}p)$ is ${(q + pt)^n}$     R1

which is a polynomial of degree $n$     R1

METHOD 2

in $n$ independent trials, it is not possible to obtain more than $n$ successes (or equivalent, eg, ${\text{P}}(X > n) = 0$)     R1

so ${a_r} = 0$ for $r > n$     R1

[2 marks]

let $Y = {X_1} + {X_2}$

${G_Y}(t) = ({q_1} + {p_1}t)({q_2} + {p_2}t)$     A1

${G_Y}(t)$ has degree two, so if $Y$ is binomial then

$Y \sim {\text{B}}(2,{\text{ }}p)$ for some $p$     R1

${(q + pt)^2} = ({q_1} + {p_1}t)({q_2} + {p_2}t)$     A1

Note:     The $LHS$ could be seen as ${q^2} + 2pqt + {p^2}{t^2}$.

METHOD 1

by considering the roots of both sides, $\frac{{{q_1}}}{{{p_1}}} = \frac{{{q_2}}}{{{p_2}}}$     M1

$\frac{{1 - {p_1}}}{{{p_1}}} = \frac{{1 - {p_2}}}{{{p_2}}}$     A1

so ${p_1} = {p_2}$     AG

METHOD 2

equating coefficients,

${p_1}{p_2} = {p^2},{\text{ }}{q_1}{q_2} = {q^2}{\text{ or }}(1 - {p_1})(1 - {p_2}) = {(1 - p)^2}$     M1

expanding,

${p_1} + {p_2} = 2p$ so ${p_1},{\text{ }}{p_2}$ are the roots of ${x^2} - 2px + {p^2} = 0$     A1

so ${p_1} = {p_2}$     AG

[5 marks]

Total [11 marks]

Solutions to (a) were often disappointing with some candidates simply writing down the answer. A common error was to forget the possibility of $X$ being zero so that $G(t) = pt$ was often seen.

Explanations in (b) were often poor, again indicating a lack of ability to give a verbal explanation.

Very few complete solutions to (c) were seen with few candidates even reaching the result that $({q_1} + {p_1}t)({q_2} + {p_2}t)$ must equal ${(q + pt)^2}$ for some $p$.