9231 P42 - Nov 2025 - Q6 - 10 marks
6620
The discrete random variable \(X\) has probability generating function \(G_X(t)\) given by
\(G_X(t)=\frac{t}{(3-2t)^2}\).
(a) Find \(\mathrm{E}(X)\) and \(\operatorname{Var}(X)\).
The discrete random variable \(Y\) has probability generating function \(G_Y(t)\) given by
\(G_Y(t)=\frac{t^2}{(3-2t)^2}\).
The random variable \(Z\) is the sum of the random variables \(X\) and \(Y\).
(b) Assuming \(X\) and \(Y\) are independent, find \(\mathrm{P}(Z\gt 4)\).
Solution
Answer: (a) \(\mathrm{E}(X)=5\), \(\operatorname{Var}(X)=12\).
(b) \(\mathrm{P}(Z\gt 4)=\frac{232}{243}\).
For a probability generating function, \(\mathrm{E}(X)=G_X'(1)\) and \(\operatorname{Var}(X)=G_X''(1)+G_X'(1)-\{G_X'(1)\}^2\).
Given \(G_X(t)=\frac{t}{(3-2t)^2}=t(3-2t)^{-2}\), differentiate:
\(G_X'(t)=(3-2t)^{-2}+4t(3-2t)^{-3}=\frac{3+2t}{(3-2t)^3}\).
So \(\mathrm{E}(X)=G_X'(1)=\frac{3+2}{1^3}=5\).
Differentiate again:
\(G_X''(t)=2(3-2t)^{-3}+6(3+2t)(3-2t)^{-4}=\frac{24+8t}{(3-2t)^4}\).
Thus \(G_X''(1)=32\), and hence
\(\operatorname{Var}(X)=32+5-5^2=12\).
Since \(X\) and \(Y\) are independent, the probability generating function of \(Z=X+Y\) is the product of the two probability generating functions:
\(G_Z(t)=G_X(t)G_Y(t)=\frac{t}{(3-2t)^2}\cdot\frac{t^2}{(3-2t)^2}=\frac{t^3}{(3-2t)^4}\).
Rewrite this in a form suitable for expansion:
\(G_Z(t)=\frac{t^3}{81}\left(1-\frac{2t}{3}\right)^{-4}\).
Using \((1-x)^{-4}=1+4x+10x^2+\cdots\), the first terms are
\(G_Z(t)=\frac{t^3}{81}\left(1+4\left(\frac{2t}{3}\right)+\cdots\right)=\frac{t^3}{81}+\frac{8t^4}{243}+\cdots\).
Therefore \(\mathrm{P}(Z=3)=\frac{1}{81}\) and \(\mathrm{P}(Z=4)=\frac{8}{243}\).
The smallest possible value of \(Z\) is \(3\), so
\(\mathrm{P}(Z\gt 4)=1-\mathrm{P}(Z=3)-\mathrm{P}(Z=4)\).
Hence
\(\mathrm{P}(Z\gt 4)=1-\frac{1}{81}-\frac{8}{243}=1-\frac{3}{243}-\frac{8}{243}=\frac{232}{243}\).
