Find \(F(u)\), the cumulative distribution function of \(U\), for \(u = 1,{\text{ }}2,{\text{ }}3 \ldots \)

Hence, or otherwise, find the value of \(P(U > 20)\).

Prove that the probability generating function of \(U\) is given by \({G_u}(t) = \frac{t}{{4 - 3t}}\).

Given that \({U_i} \sim {\text{Geo}}\left( {\frac{1}{4}} \right),{\text{ }}i = 1,{\text{ }}2,{\text{ }}3\), and that \(V = {U_1} + {U_2} + {U_3}\), find

(i)     \({\text{E}}(V)\);

(ii)     \({\text{Var}}(V)\);

(iii)     \({G_v}(t)\), the probability generating function of \(V\).

A third random variable \(W\), has probability generating function \({G_w}(t) = \frac{1}{{{{(4 - 3t)}^3}}}\).

By differentiating \({G_w}(t)\), find \({\text{E}}(W)\).

A third random variable \(W\), has probability generating function \({G_w}(t) = \frac{1}{{{{(4 - 3t)}^3}}}\).

Prove that \(V = W + 3\).

METHOD 1

\({\text{P}}(U = u) = \frac{1}{4}{\left( {\frac{3}{4}} \right)^{u - 1}}\)     (M1)

\(F(u) = {\text{P}}(U \le u) = \sum\limits_{r = 1}^u {\frac{1}{4}{{\left( {\frac{3}{4}} \right)}^{r - 1}}\;\;\;} \)(or equivalent)

\( = \frac{{\frac{1}{4}\left( {1 - {{\left( {\frac{3}{4}} \right)}^u}} \right)}}{{1 - \frac{3}{4}}}\)     (M1)

\( = 1 - {\left( {\frac{3}{4}} \right)^u}\)     A1

METHOD 2

\({\text{P}}(U \le u) = 1 - {\text{P}}(U > u)\)     (M1)

\({\text{P}}(U > u) = \) probability of \(u\) consecutive failures     (M1)

\({\text{P}}(U \le u) = 1 - {\left( {\frac{3}{4}} \right)^u}\)     A1

[3 marks]

\({\text{P}}(U > 20) = 1 - {\text{P}}(U \le 20)\)     (M1)

\( = {\left( {\frac{3}{4}} \right)^{20}}\;\;\;( = 0.00317)\)     A1

[2 marks]

\({G_U}(t) = \sum\limits_{r = 1}^\infty  {\frac{1}{4}{{\left( {\frac{3}{4}} \right)}^{r - 1}}{t^r}\;\;\;} \)(or equivalent)     M1A1

\( = \sum\limits_{r = 1}^\infty  {\frac{1}{3}{{\left( {\frac{3}{4}t} \right)}^r}} \)     (M1)

\( = \frac{{\frac{1}{3}\left( {\frac{3}{4}t} \right)}}{{1 - \frac{3}{4}t}}\;\;\;\left( { = \frac{{\frac{1}{4}t}}{{1 - \frac{3}{4}t}}} \right)\)     A1

\( = \frac{t}{{4 - 3t}}\)     AG

[4 marks]

(i)     \(E(U) = \frac{1}{{\frac{1}{4}}} = 4\)     (A1)

\(E({U_1} + {U_2} + {U_3}{\text{)}} = 4 + 4 + 4 = 12\)     A1

(ii)     \({\text{Var}}(U) = \frac{{\frac{3}{4}}}{{{{\left( {\frac{1}{4}} \right)}^2}}}=12\)     A1

\({\text{Var(}}{U_1} + {U_2} + {U_3}) = 12 + 12 + 12 = 36\)     A1

(iii)     \({G_v}(t) = {\left( {{G_U}(t)} \right)^3}\)     (M1)

\( = {\left( {\frac{t}{{4 - 3t}}} \right)^3}\)     A1

[6 marks]

\({G_W}^\prime (t) =  - 3{(4 - 3t)^{ - 4}}( - 3)\;\;\;\left( { = \frac{9}{{{{(4 - 3t)}^4}}}} \right)\)     (M1)(A1)

\(E(W) = {G_W}^\prime (1) = 9\)     (M1)A1

Note:     Allow the use of the calculator to perform the differentiation.

[4 marks]

EITHER

probability generating function of the constant 3 is \({t^3}\)     A1

OR

\({G_{W - 3}}(t) = E({t^{W + 3}}) = E({t^W})E({t^3})\)     A1

THEN

\(W + 3\) has generating function \({G_{W + 3}} = \frac{1}{{{{(4 - 3t)}^3}}} \times {t^3} = {G_V}(t)\)     M1

as the generating functions are the same \(V = W + 3\)     R1AG

[3 marks]

Total [22 marks]