Answer:

1. (a) \(\mathrm{G}_{X}(t)=\frac{2}{5}t+\frac{3}{5}t^{2}\).
2. (b) \(\mathrm{G}_{Y}(t)=\frac{1}{625}t^{4}(2+3t)^{4}\), so \(a=\frac{1}{625}\), \(b=2\), \(c=3\), \(m=4\), \(n=4\).
3. (c) \(\mathrm{P}(Y=6)=\frac{216}{625}\).
4. (d) \(\operatorname{Var}(Y)=\frac{24}{25}\).

1. (a)

   For a probability generating function, \(\mathrm{G}_{X}(t)=\mathrm{E}(t^X)\). Since \(X\) takes the values 1 and 2,

   \(\mathrm{G}_{X}(t)=\frac{2}{5}t^1+\frac{3}{5}t^2=\frac{2}{5}t+\frac{3}{5}t^2\).

2. (b)

   Let \(Y=X_1+X_2+X_3+X_4\), where the four observations are independent and each has the same distribution as \(X\). The probability generating function of a sum of independent random variables is the product of their probability generating functions, so

   \(\mathrm{G}_{Y}(t)=\left(\mathrm{G}_{X}(t)\right)^4=\left(\frac{2}{5}t+\frac{3}{5}t^2\right)^4\).

   Factorising inside the bracket gives

   \(\mathrm{G}_{Y}(t)=\left(t\left(\frac{2}{5}+\frac{3}{5}t\right)\right)^4=t^4\left(\frac{2+3t}{5}\right)^4=\frac{1}{625}t^4(2+3t)^4\).

3. (c)

   The coefficient of \(t^6\) in \(\mathrm{G}_{Y}(t)\) is \(\mathrm{P}(Y=6)\).

   From \(\mathrm{G}_{Y}(t)=\frac{1}{625}t^4(2+3t)^4\), we need the coefficient of \(t^2\) in \((2+3t)^4\).

   Using the binomial expansion, the \(t^2\) term is

   \(\binom{4}{2}2^2(3t)^2=6\cdot4\cdot9t^2=216t^2\).

   Therefore \(\mathrm{P}(Y=6)=\frac{216}{625}\).

4. (d)

   Since \(Y\) is the sum of four independent observations of \(X\),

   \(\operatorname{Var}(Y)=4\operatorname{Var}(X)\).

   Now

   \(\mathrm{E}(X)=1\cdot\frac{2}{5}+2\cdot\frac{3}{5}=\frac{2}{5}+\frac{6}{5}=\frac{8}{5}\),

   and

   \(\mathrm{E}(X^2)=1^2\cdot\frac{2}{5}+2^2\cdot\frac{3}{5}=\frac{2}{5}+\frac{12}{5}=\frac{14}{5}\).

   Hence

   \(\operatorname{Var}(X)=\mathrm{E}(X^2)-\left(\mathrm{E}(X)\right)^2=\frac{14}{5}-\left(\frac{8}{5}\right)^2=\frac{70}{25}-\frac{64}{25}=\frac{6}{25}\).

   Therefore

   \(\operatorname{Var}(Y)=4\cdot\frac{6}{25}=\frac{24}{25}\).