Show that the probability distribution of \(Y\) is given by \({\text{P}}(Y = y) = {{\text{e}}^{ - y}}(1 - {{\text{e}}^{ - 1}}),{\text{ }}y \in \mathbb{N}\).

(i)     Show that \(G(t)\), the probability generating function of \(Y\), is given by \(G(t) = \frac{{1 - {{\text{e}}^{ - 1}}}}{{1 - {{\text{e}}^{ - 1}}t}}\).

(ii)     Hence determine the value of \({\text{E}}(Y)\) correct to three significant figures.

\({\text{P}}(Y = y) = \int_y^{y + 1} {{{\text{e}}^{ - x}}{\text{d}}x} \)    M1A1

\( = {[ - {{\text{e}}^{ - x}}]^{y + 1}}y\)    A1

\( =  - {{\text{e}}^{ - (y + 1)}} + {{\text{e}}^{ - y}}\)    A1

\( = {{\text{e}}^{ - y}}(1 - {{\text{e}}^{ - 1}})\)    AG

[4 marks]

(i)     attempt to use \(G(t) = \sum {{\text{P}}(Y = y){t^y}} \)     (M1)

\( = \sum\limits_{y = 0}^\infty  {{{\text{e}}^{ - y}}(1 - {{\text{e}}^{ - 1}}){t^y}} \)    A1

Note:     Accept a listing of terms without the use of \(\Sigma \).

this is an infinite geometric series with first term \(1 - {{\text{e}}^{ - 1}}\) and common ratio \({{\text{e}}^{ - 1}}t\)     M1

\(G(t) = \frac{{1 - {{\text{e}}^{ - 1}}}}{{1 - {{\text{e}}^{ - 1}}t}}\)    AG

(ii)     \({\text{E}}(Y) = G'(1)\)     M1

\(G'(t) = \frac{{1 - {{\text{e}}^{ - 1}}}}{{{{(1 - {{\text{e}}^{ - 1}}t)}^2}}} \times {{\text{e}}^{ - 1}}\)     (M1)(A1)

\({\text{E}}(Y) = \frac{{{{\text{e}}^{ - 1}}}}{{(1 - {{\text{e}}^{ - 1}})}}\)    (A1)

\( = 0.582\)    A1

Note:     Allow the use of GDC to determine \(G'(1)\).

[8 marks]

In (a), it was disappointing to find that very few candidates realised that \({\text{P}}(Y = y)\) could be found by integrating \(f(x)\) from \(y\) to \(y + 1\). Candidates who simply integrated \(f(x)\) to find the cumulative distribution function of \(X\) were given no credit unless they attempted to use their result to find the probability distribution of \(Y\).

Solutions to (b)(i) were generally good although marks were lost due to not including the \(y = 0\) term.

Part (b)(ii) was also well answered in general with the majority of candidates using the GDC to evaluate \(G'(1)\).

Candidates who tried to differentiate \(G(t)\) algebraically often made errors.