# IB DP Maths Topic 7.1 Geometric distribution HL Paper 3

## Question

Anna has a fair cubical die with the numbers 1, 2, 3, 4, 5, 6 respectively on the six faces. When she tosses it, the score is defined as the number on the uppermost face. One day, she decides to toss the die repeatedly until all the possible scores have occurred at least once.

(a)     Having thrown the die once, she lets $${X_2}$$ denote the number of additional throws required to obtain a different number from the one obtained on the first throw. State the distribution of $${X_2}$$ and hence find $${\text{E}}({X_2})$$ .

(b)     She then lets $${X_3}$$ denote the number of additional throws required to obtain a different number from the two numbers already obtained. State the distribution of $${X_3}$$ and hence find $${\text{E}}({X_3})$$ .

(c)     By continuing the process, show that the expected number of tosses needed to obtain all six possible scores is 14.7.

## Markscheme

(a)     $${X_2}$$ is a geometric random variable     A1

with $$p = \frac{5}{6}.$$     A1

Therefore $${\text{E}}({X_2}) = \frac{6}{5}.$$     A1

[3 marks]

(b)     $${X_3}$$ is a geometric random variable with $$p = \frac{4}{6}.$$     A1

Therefore $${\text{E}}({X_3}) = \frac{6}{4}.$$     A1

[2 marks]

(c)     $${\text{E}}({X_4}) = \frac{6}{3},{\text{ E}}({X_5}) = \frac{6}{2},{\text{ E}}({X_6}) = \frac{6}{1}$$     A1A1A1

$${\text{E}}({X_1}) = 1\,\,\,\,\,{\text{(or }}{X_1} = 1)$$     A1

Expected number of tosses $$\sum\limits_{n = 1}^6 {{\text{E}}({X_n})}$$     M1

$$= 14.7$$     AG

[5 marks]

Total [10 marks]

## Examiners report

Many candidates were unable even to start this question although those who did often made substantial progress.

## Question

In a game there are n players, where $$n > 2$$ . Each player has a disc, one side of which is red and one side blue. When thrown, the disc is equally likely to show red or blue. All players throw their discs simultaneously. A player wins if his disc shows a different colour from all the other discs. Players throw repeatedly until one player wins.

Let X be the number of throws each player makes, up to and including the one on which the game is won.

(a)     State the distribution of X .

(b)     Find $${\text{P}}(X = x)$$ in terms of n and x .

(c)     Find $${\text{E}}(X)$$ in terms of n .

(d)     Given that n = 7 , find the least number, k , such that $${\text{P}}(X \leqslant k) > 0.5$$ .

## Markscheme

(a)     geometric distribution     A1

[1 mark]

(b)     let R be the event throwing the disc and it landing on red and

let B be the event throwing the disc and it landing on blue

$${\text{P}}(X = 1) = p = {\text{P}}\left( {1B{\text{ and }}(n – 1)R{\text{ or }}1R{\text{ and }}(n – 1)B} \right)$$     (M1)

$$= n \times \frac{1}{2} \times {\left( {\frac{1}{2}} \right)^{n – 1}} + n \times \frac{1}{2} \times {\left( {\frac{1}{2}} \right)^{n – 1}}$$     (A1)

$$= \frac{n}{{{2^{n – 1}}}}$$     A1

hence $${\text{P}}(X = x) = \frac{n}{{{2^{n – 1}}}}{\left( {1 – \frac{n}{{{2^{n – 1}}}}} \right)^{x – 1}},{\text{ }}(x \geqslant 1)$$     A1

Notes: $$x \geqslant 1$$ not required for final A1.

Allow FT for final A1.

[4 marks]

(c)     $${\text{E}}(X) = \frac{1}{p}$$

$$= \frac{{{2^{n – 1}}}}{n}$$     A1

[1 mark]

(d)     when $$n = 7$$ , $${\text{P}}(X = x) = {\left( {1 – \frac{7}{{64}}} \right)^{x – 1}} \times \frac{7}{{64}}$$     (M1)

$$= \frac{7}{{64}} \times {\left( {\frac{{57}}{{64}}} \right)^{x – 1}}$$

$${\text{P}}(X \leqslant k) = \sum\limits_{x = 1}^k {\frac{7}{{64}} \times {{\left( {\frac{{57}}{{64}}} \right)}^{x – 1}}}$$     (M1)(A1)

$$\Rightarrow \frac{7}{{64}} \times \frac{{1 – {{\left( {\frac{{57}}{{64}}} \right)}^k}}}{{1 – \frac{{57}}{{64}}}} > 0.5$$     (M1)(A1)

$$\Rightarrow 1 – {\left( {\frac{{57}}{{64}}} \right)^k} > 0.5$$

$$\Rightarrow {\left( {\frac{{57}}{{64}}} \right)^k} < 0.5$$

$$\Rightarrow k > \frac{{\log 0.5}}{{\log \frac{{57}}{{64}}}}$$     (M1)

$$\Rightarrow k > 5.98$$     (A1)

$$\Rightarrow k = 6$$     A1

Note: Tabular and other GDC methods are acceptable.

[8 marks]

Total [14 marks]

## Examiners report

This question was found difficult by the majority of candidates and few fully correct answers were seen. Few candidates were able to find $${\text{P}}(X = x)$$ in terms of n and x and many did not realise that the last part of the question required them to find the sum of a series. However, better candidates received over 75% of the marks because the answers could be followed through.

## Question

The random variable X has a geometric distribution with parameter p .

Show that $${\text{P}}(X \leqslant n) = 1 – {(1 – p)^n},{\text{ }}n \in {\mathbb{Z}^ + }$$ .


a.

Deduce an expression for $${\text{P}}(m < X \leqslant n)\,,{\text{ }}m\,,{\text{ }}n \in {\mathbb{Z}^ + }$$ and m < n .


b.

Given that p = 0.2, find the least value of n for which $${\text{P}}(1 < X \leqslant n) > 0.5\,,{\text{ }}n \in {\mathbb{Z}^ + }$$ .


c.

## Markscheme

$${\text{P}}(X \leqslant n) = \sum\limits_{{\text{i}} = 1}^n {{\text{P}}(X = {\text{i}}) = \sum\limits_{{\text{i}} = 1}^n {p{q^{{\text{i}} – 1}}} }$$     M1A1

$$= p\frac{{1 – {q^n}}}{{1 – q}}$$     A1

$$= 1 – {(1 – p)^n}$$     AG

[3 marks]

a.

$${(1 – p)^m} – {(1 – p)^n}$$     A1

[1 mark]

b.

attempt to solve $$0.8 – {(0.8)^n} > 0.5$$     M1

obtain n = 6     A1

[2 marks]

c.

## Examiners report

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

a.

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

b.

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

c.

## Question

Jenny and her Dad frequently play a board game. Before she can start Jenny has to throw a “six” on an ordinary six-sided dice. Let the random variable X denote the number of times Jenny has to throw the dice in total until she obtains her first “six”.

If the dice is fair, write down the distribution of X , including the value of any parameter(s).


a.

Write down E(X ) for the distribution in part (a).


b.

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Write down the distribution of Y , including the value of any parameter(s).


d.

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Find the value of y such that $${\text{P}}(Y = y) = \frac{1}{{36}}$$.


e.

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Find $${\text{P}}(Y \leqslant 6)$$ .


f.

## Markscheme

$$X \sim {\text{Geo}}\left( {\frac{1}{6}} \right){\text{ or NB}}\left( {1,\frac{1}{6}} \right)$$     A1

[1 mark]

a.

$${\text{E}}(X) = 6$$     A1

[1 mark]

b.

Y is $${\text{NB}}\left( {2,\frac{1}{6}} \right)$$     A1

[1 mark]

d.

$${\text{P}}(Y = y) = \frac{1}{{36}}{\text{ gives }}y = 2$$     A1

(as all other probabilities would have a factor of 5 in the numerator)

[1 mark]

e.

$${\text{P}}(Y \leqslant 6) = {\left( {\frac{1}{6}} \right)^2} + 2\left( {\frac{5}{6}} \right){\left( {\frac{1}{6}} \right)^2} + 3{\left( {\frac{5}{6}} \right)^2}{\left( {\frac{1}{6}} \right)^2} + 4{\left( {\frac{5}{6}} \right)^3}{\left( {\frac{1}{6}} \right)^2} + 5{\left( {\frac{5}{6}} \right)^4}{\left( {\frac{1}{6}} \right)^2}$$     (M1)

$$= 0.263$$     A1

[2 marks]

f.

## Examiners report

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

a.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

b.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

d.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

e.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

f.

## Question

Consider the random variable $$X \sim {\text{Geo}}(p)$$.

(a)     State $${\text{P}}(X < 4)$$.

(b)     Show that the probability generating function for X is given by $${G_X}(t) = \frac{{pt}}{{1 – qt}}$$, where $$q = 1 – p$$.

Let the random variable $$Y = 2X$$.

(c)     (i)     Show that the probability generating function for Y is given by $${G_Y}(t) = {G_X}({t^2})$$.

(ii)     By considering $${G’_Y}(1)$$, show that $${\text{E}}(Y) = 2{\text{E}}(X)$$.

Let the random variable $$W = 2X + 1$$.

(d)     (i)     Find the probability generating function for W in terms of the probability generating function of Y.

(ii)     Hence, show that $${\text{E}}(W) = 2{\text{E}}(X) + 1$$.

## Markscheme

(a)     use of $${\text{P}}(X = n) = p{q^{n – 1}}{\text{ }}(q = 1 – p)$$     (M1)

$${\text{P}}(X < 4) = p + pq + p{q^2}{\text{ }}\left( { = 1 – {q^3}} \right){\text{ }}\left( { = 1 – {{(1 – p)}^3}} \right){\text{ }}( = 3p – 3{p^2} + {p^3})$$     A1

[2 marks]

(b)   $${G_X}(t) = {\text{P}}(X = 1)t + {\text{P}}(X = 2){t^2} + \ldots$$     (M1)

$$= pt + pq{t^2} + p{q^2}{t^3} + \ldots$$     A1

summing an infinite geometric series     M1

$$= \frac{{pt}}{{1 – qt}}$$     AG

[3 marks]

(c)     (i)     EITHER

$${G_Y}(t) = {\text{P}}(Y = 1)t + {\text{P}}(Y = 2){t^2} + \ldots$$     A1

$$= 0 \times t + {\text{P}}(X = 1){t^2} + 0 \times {t^3} + {\text{P}}(X = 2){t^4} + \ldots$$     M1A1

$$= {G_X}({t^2})$$     AG

OR

$${G_Y}(t) = E({t^Y}) = E({t^{2X}})$$     M1A1

$$= E\left( {{{({t^2})}^X}} \right)$$     A1

$$= {G_X}({t^2})$$     AG

(ii)     $${\text{E}}(Y) = {G’_Y}(1)$$     A1

EITHER

$$= 2t{G’_X}({t^2})$$ evaluated at $$t = 1$$     M1A1

$$= 2{\text{E}}(X)$$     AG

OR

$$= \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{p{t^2}}}{{(1 – q{t^2})}}} \right) = \frac{{2pt(1 – q{t^2}) + 2pq{t^3}}}{{{{(1 – q{t^2})}^2}}}$$ evaluated at $$t = 1$$     A1

$$= 2 \times \frac{{p(1 – qt) + pqt}}{{{{(1 – qt)}^2}}}$$ evaluated at $$t = 1{\text{ (or }}\frac{2}{p})$$     A1

$$= 2{\text{E}}(X)$$     AG

[6 marks]

(d)     (i)     $${G_W}(t) = t{G_Y}(t)$$ (or equivalent)     A2

(ii)     attempt to evaluate $${G’_W}(t)$$     M1

EITHER

obtain $$1 \times {G_Y}(t) + t \times {G’_Y}(t)$$     A1

substitute $$t = 1$$ to obtain $$1 \times 1 + 1 \times {G’_Y}(1)$$     A1

OR

$$= \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{p{t^3}}}{{(1 – q{t^2})}}} \right) = \frac{{3p{t^2}(1 – q{t^2}) + 2pq{t^4}}}{{{{(1 – q{t^2})}^2}}}$$     A1

substitute $$t = 1$$ to obtain $$1 + \frac{2}{p}$$     A1

$$= 1 + 2{\text{E}}(X)$$     AG

[5 marks]

Total [16 marks]

[N/A]

## Question

A discrete random variable $$U$$ follows a geometric distribution with $$p = \frac{1}{4}$$.

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


a.

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


b.

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


c.

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 $$V$$.


d.

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

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


e.

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

Prove that $$V = W + 3$$.


f.

## Markscheme

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]

a.

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

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

[2 marks]

b.

$${G_U}(t) = \sum\limits_{r = 1}^\infty {\frac{1}{4}{{\left( {\frac{3}{4}} \right)}^{r – 1}}{t^r}\;\;\;}$$(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]

c.

(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]

d.

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

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

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

[4 marks]

e.

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]

f.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

[N/A]

f.

## Question

The continuous random variable $$X$$ has probability density function

$f(x) = \left\{ {\begin{array}{*{20}{c}} {{{\text{e}}^{ – x}}}&{x \geqslant 0} \\ 0&{x < 0} \end{array}.} \right.$

The discrete random variable $$Y$$ is defined as the integer part of $$X$$, that is the largest integer less than or equal to $$X$$.

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


a.

(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.


b.

## Markscheme

$${\text{P}}(Y = y) = \int_y^{y + 1} {{{\text{e}}^{ – x}}{\text{d}}x}$$    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]

a.

(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]

b.

## Examiners report

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 $$Y$$.

a.

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.

b.

## Question

The continuous random variable $$X$$ has probability density function

$f(x) = \left\{ {\begin{array}{*{20}{c}} {{{\text{e}}^{ – x}}}&{x \geqslant 0} \\ 0&{x < 0} \end{array}.} \right.$

The discrete random variable $$Y$$ is defined as the integer part of $$X$$, that is the largest integer less than or equal to $$X$$.

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


a.

(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.


b.

## Markscheme

$${\text{P}}(Y = y) = \int_y^{y + 1} {{{\text{e}}^{ – x}}{\text{d}}x}$$    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]

a.

(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]

b.

## Examiners report

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 $$Y$$.

a.

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.

b.

## Question

Two independent discrete random variables $$X$$ and $$Y$$ have probability generating functions $$G(t)$$ and $$H(t)$$ respectively. Let $$Z = X + Y$$ have probability generating function $$J(t)$$.

Write down an expression for $$J(t)$$ in terms of $$G(t)$$ and $$H(t)$$.


a.

By differentiating $$J(t)$$, prove that

(i)     $${\text{E}}(Z) = {\text{E}}(X) + {\text{E}}(Y)$$;

(ii)     $${\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)$$.


b.

## Markscheme

$$J(t) = G(t)H(t)$$    A1

[1 mark]

a.

(i)     $$J'(t) = G'(t)H(t) + G(t)H'(t)$$     M1A1

$$J'(1) = G'(1)H(1) + G(1)H'(1)$$    M1

$$J'(1) = G'(1) + H'(1)$$    A1

so $$E(Z) = E(X) + E(Y)$$     AG

(ii)     $$J”(t) = G”(t)H(t) + G'(t)H'(t) + G'(t)H'(t) + G(t)H”(t)$$     M1A1

$$J”(1) = G”(1)H(1) + 2G'(1)H'(1) + G(1)H”(1)$$

$$= G”(1) + 2G'(1)H'(1) + H”(1)$$    A1

$${\text{Var}}(Z) = J”(1) + J'(1) – {\left( {J'(1)} \right)^2}$$    M1

$$= G”(1) + 2G'(1)H'(1) + H”(1) + G'(1) + H'(1) – {\left( {G'(1) + H'(1)} \right)^2}$$    A1

$$= G”(1) + G'(1) – {\left( {G'(1)} \right)^2} + H”(1) + H'(1) – {\left( {H'(1)} \right)^2}$$    A1

so $${\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)$$     AG

Note: If addition is wrongly used instead of multiplication in (a) it is inappropriate to give FT apart from the second M marks in each part, as the working is too simple.

[10 marks]

b.

[N/A]

a.

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b.

## Question

A continuous random variable $$T$$ has a probability density function defined by

$$f(t) = \left\{ {\begin{array}{*{20}{c}} {\frac{{t(4 – {t^2})}}{4}}&{0 \leqslant t \leqslant 2} \\ {0,}&{{\text{otherwise}}} \end{array}} \right.$$.

Find the cumulative distribution function $$F(t)$$, for $$0 \leqslant t \leqslant 2$$.


a.

Sketch the graph of $$F(t)$$ for $$0 \leqslant t \leqslant 2$$, clearly indicating the coordinates of the endpoints.


b.i.

Given that $$P(T < a) = 0.75$$, find the value of $$a$$.


b.ii.

## Markscheme

$$F(t) = \int_0^t {\left( {x – \frac{{{x^3}}}{4}} \right){\text{d}}x{\text{ }}\left( { = \int_0^t {\frac{{x(4 – {x^2})}}{4}{\text{d}}x} } \right)}$$     M1

$$= \left[ {\frac{{{x^2}}}{2} – \frac{{{x^4}}}{{16}}} \right]_0^t{\text{ }}\left( { = \left[ {\frac{{{x^2}(8 – {x^2})}}{{16}}} \right]_0^t} \right){\text{ }}\left( { = \left[ {\frac{{ – 4 – {x^2}{)^2}}}{{16}}} \right]_0^t} \right)$$     A1

$$= \frac{{{t^2}}}{2} – \frac{{{t^4}}}{{16}}{\text{ }}\left( { = \frac{{{t^2}(8 – {t^2})}}{{16}}} \right){\text{ }}\left( { = 1 – \frac{{{{(4 – {t^2})}^2}}}{{16}}} \right)$$     A1

Note:     Condone integration involving $$t$$ only.

Note:     Award M1A0A0 for integration without limits eg, $$\int {\frac{{t(4 – {t^2})}}{4}{\text{d}}t = \frac{{{t^2}}}{2} – \frac{{{t^4}}}{{16}}}$$ or equivalent.

Note:     But allow integration $$+$$ $$C$$ then showing $$C = 0$$ or even integration without $$C$$ if $$F(0) = 0$$ or $$F(2) = 1$$ is confirmed.

[3 marks]

a. correct shape including correct concavity     A1

clearly indicating starts at origin and ends at $$(2,{\text{ }}1)$$     A1

Note:     Condone the absence of $$(0,{\text{ }}0)$$.

Note:     Accept 2 on the $$x$$-axis and 1 on the $$y$$-axis correctly placed.

[2 marks]

b.i.

attempt to solve $$\frac{{{a^2}}}{2} – \frac{{{a^4}}}{{16}} = 0.75$$ (or equivalent) for $$a$$     (M1)

$$a = 1.41{\text{ }}( = \sqrt 2 )$$     A1

Note:     Accept any answer that rounds to 1.4.

[2 marks]

b.ii.

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

## Question

The random variable $$X$$ follows a Poisson distribution with mean $$\lambda$$. The probability generating function of $$X$$ is given by $${G_X}(t) = {{\text{e}}^{\lambda (t – 1)}}$$.

The random variable $$Y$$, independent of $$X$$, follows a Poisson distribution with mean $$\mu$$.

Find expressions for $${G’_X}(t)$$ and $${G’’_X}(t)$$.


a.i.

Hence show that $${\text{Var}}(X) = \lambda$$.


a.ii.

By considering the probability generating function, $${G_{X + Y}}(t)$$, of $$X + Y$$, show that $$X + Y$$ follows a Poisson distribution with mean $$\lambda + \mu$$.


b.

Show that $${\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 – \frac{\lambda }{{\lambda + \mu }}} \right)^{n – x}}$$, where $$n$$, $$x$$ are non-negative integers and $$n \geqslant x$$.


c.i.

Identify the probability distribution given in part (c)(i) and state its parameters.


c.ii.

## Markscheme

$${G’_X}(t) = \lambda {{\text{e}}^{\lambda (t – 1)}}$$     A1

$${G’’_X}(t) = {\lambda ^2}{{\text{e}}^{\lambda (t – 1)}}$$     A1

[2 marks]

a.i.

$${\text{Var}}(X) = {G”_X}(1) + {G’_X}(1) – {\left( {{{G’}_X}(1)} \right)^2}$$     (M1)

$${G’_X}(1) = \lambda$$ and $${G’’_X}(1) = {\lambda ^2}$$     (A1)

$${\text{Var}}(X) = {\lambda ^2} + \lambda – {\lambda ^2}$$     A1

$$= \lambda$$     AG

[3 marks]

a.ii.

$${G_{X + Y}}(t) = {{\text{e}}^{\lambda (t – 1)}} \times {{\text{e}}^{\mu (t – 1)}}$$     M1

Note:     The M1 is for knowing to multiply pgfs.

$$= {{\text{e}}^{(\lambda + \mu )(t – 1)}}$$     A1

which is the pgf for a Poisson distribution with mean $$\lambda + \mu$$     R1AG

Note:     Line 3 identifying the Poisson pgf must be seen.

[3 marks]

b.

$${\text{P}}(X = x|X + Y = n) = \frac{{{\text{P}}(X = x \cap Y = n – x)}}{{{\text{P}}(X + Y = n)}}$$     (M1)

$$= \left( {\frac{{{{\text{e}}^{ – \lambda }}{\lambda ^x}}}{{x!}}} \right)\left( {\frac{{{{\text{e}}^{ – \mu }}{\mu ^{n – x}}}}{{(n – x)!}}} \right)\left( {\frac{{n!}}{{{{\text{e}}^{ – (\lambda + \mu )}}{{(\lambda + \mu )}^n}}}} \right)$$ (or equivalent)     M1A1

$$= \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right)\frac{{{\lambda ^x}{\mu ^{n – x}}}}{{{{(\lambda + \mu )}^n}}}$$     A1

$$= \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {\frac{\mu }{{\lambda + \mu }}} \right)^{n – x}}$$     A1

leading to $${\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 – \frac{\lambda }{{\lambda + \mu }}} \right)^{n – x}}$$     AG

[5 marks]

c.i.

$${\text{B}}\left( {n,{\text{ }}\frac{\lambda }{{\lambda + \mu }}} \right)$$     A1A1

Note:     Award A1 for stating binomial and A1 for stating correct parameters.

[2 marks]

c.ii.

[N/A]

a.i.

[N/A]

a.ii.

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b.

[N/A]

c.i.

[N/A]

c.ii.

## Question

Consider an unbiased tetrahedral (four-sided) die with faces labelled 1, 2, 3 and 4 respectively.

The random variable X represents the number of throws required to obtain a 1.

State the distribution of X.


a.

Show that the probability generating function, $$G\left( t \right)$$, for X is given by $$G\left( t \right) = \frac{t}{{4 – 3t}}$$.


b.

Find $$G’\left( t \right)$$.


c.

Determine the mean number of throws required to obtain a 1.


d.

## Markscheme

X is geometric (or negative binomial)      A1

[1 mark]

a.

$$G\left( t \right) = \frac{1}{4}t + \frac{1}{4}\left( {\frac{3}{4}} \right){t^2} + \frac{1}{4}{\left( {\frac{3}{4}} \right)^2}{t^3} + \ldots$$     M1A1

recognition of GP $$\left( {{u_1} = \frac{1}{4}t,\,\,r = \frac{3}{4}t} \right)$$     (M1)

$$= \frac{{\frac{1}{4}t}}{{1 – \frac{3}{4}t}}$$     A1

leading to $$G\left( t \right) = \frac{t}{{4 – 3t}}$$     AG

[4 marks]

b.

attempt to use product or quotient rule      M1

$$G’\left( t \right) = \frac{4}{{{{\left( {4 – 3t} \right)}^2}}}$$     A1

[2 marks]

c.

4      A1

Note: Award A1FT to a candidate that correctly calculates the value of $$G’\left( 1 \right)$$ from their $$G’\left( t \right)$$.

[1 mark]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.