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]