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 \(p\).