# IB DP Maths Topic 7.1 Negative binomial distribution HL Paper 3

## Question

(a)     The heating in a residential school is to be increased on the third frosty day during the term. If the probability that a day will be frosty is 0.09, what is the probability that the heating is increased on the $${25^{{\text{th}}}}$$ day of the term?

(b)     On which day is the heating most likely to be increased?

## Markscheme

(a)     the distribution is NB(3, 0.09)     (M1)(A1)

the probability is $$\left( {\begin{array}{*{20}{c}} {24} \\ 2 \end{array}} \right){0.91^{22}} \times {0.09^3} = 0.0253$$     (M1)(A1)A1

[5 marks]

(b)     P(Heating increased on $${n^{{\text{th}}}}$$ day)

$$\left( {\begin{array}{*{20}{c}} {n – 1} \\ 2 \end{array}} \right){0.91^{n – 3}} \times {0.09^3}$$     (M1)(A1)(A1)

by trial and error n = 23 gives the maximum probability     (M1)A3

(neighbouring values: 0.02551 (n = 22) ; 0.02554 (n = 23) ; 0.02545 (n = 24) )

[7 marks]

Total [12 marks]

## Examiners report

Most candidates understood the context of this question, and the negative binomial distribution was usually applied, albeit occasionally with incorrect parameters. Good solutions were seen to part(b), using lists in their GDC or trial and error.

## Question

The random variable X has the negative binomial distribution NB(5, p), where p < 0.5, and $${\text{P}}(X = 10) = 0.05$$. By first finding the value of p, find the value of $${\text{P}}(X = 11)$$.

## Markscheme

$${\text{P}}(X = 10) = \left( {\begin{array}{*{20}{c}} 9 \\ 4 \end{array}} \right){p^5}{(1 – p)^5}$$     (= 0.05)     (M1)A1A1

Note: First A1 is for the binomial coefficient. Second A1 is for the rest.

solving by any method, $$p = 0.297 \ldots$$     A4

Notes: Award A2 for anything which rounds to 0.703.

Do not apply any AP at this stage.

$${\text{P}}(X = 10) = \left( {\begin{array}{*{20}{c}} {10} \\ 4 \end{array}} \right) \times {(0.297…)^5} \times {(1 – 0.297…)^6}$$     (M1)A1

= 0.0586     A1

Note: Allow follow through for incorrect p-values.

[10 marks]

## Examiners report

Questions on these discrete distributions have not been generally well answered in the past and it was pleasing to note that many candidates submitted a reasonably good solution to this question. In (b), the determination of the value of p was often successful using a variety of methods including solving the equation $$p(1 – p) = {(0.000396{\text{ }} \ldots )^{1/5}}$$, graph plotting or using SOLVER on the GDC or even expanding the equation into a $${10^{{\text{th}}}}$$ degree polynomial and solving that. Solutions to this particular question exceeded expectations.

## Question

As soon as Sarah misses a total of 4 lessons at her school an email is sent to her parents. The probability that she misses any particular lesson is constant with a value of $$\frac{1}{3}$$. Her decision to attend a lesson is independent of her previous decisions.

(a)     Find the probability that an email is sent to Sarah’s parents after the $${8^{{\text{th}}}}$$ lesson that Sarah was scheduled to attend.

(b)     If an email is sent to Sarah’s parents after the $${X^{{\text{th}}}}$$ lesson that she was scheduled to attend, find $${\text{E}}(X)$$.

(c)     If after 6 of Sarah’s scheduled lessons we are told that she has missed exactly 2 lessons, find the probability that an email is sent to her parents after a total of 12 scheduled lessons.

(d)     If we know that an email was sent to Sarah’s parents immediately after her $${6^{{\text{th}}}}$$ scheduled lesson, find the probability that Sarah missed her $${2^{{\text{nd}}}}$$ scheduled lesson.

## Markscheme

(a)     we are dealing with the Negative Binomial distribution: $${\text{NB}}\left( {4,\frac{1}{3}} \right)$$     (M1)

let X be the number of scheduled lessons before the email is sent

$${\text{P}}(X = 8) = \left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^4} = 0.0854$$     (M1)A1

[3 marks]

(b)     $${\text{E}}(X) = \frac{r}{p} = \frac{4}{{\frac{1}{3}}} = 12$$     (M1)A1

[2 marks]

(c)     we are asking for 2 missed lessons in the second 6 lessons, with the last lesson missed so this is $${\text{NB}}\left( {2,\frac{1}{3}} \right)$$     (M1)

$${\text{P}}(X = 6) = \left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^2} = 0.110$$     (M1)A1

Note: Accept solutions laid out in terms of conditional probabilities.

[3 marks]

(d)     EITHER

We know that she missed the $${6^{{\text{th}}}}$$ lesson so she must have missed 3 from the first 5 lessons. All are equally likely so the probability that she missed the $${2^{{\text{nd}}}}$$ lesson is $$\frac{3}{5}$$.     R1A1

OR

require $${\text{P(missed }}{{\text{2}}^{{\text{nd}}}}|X = 6) = \frac{{{\text{P(missed }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}X = 6)}}{{{\text{P}}(X = 6)}}$$     R1

$${\text{P(missed }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}X = 6) = {\text{P(missed }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}{{\text{6}}^{{\text{th}}}}{\text{ and 2 of remaining 4)}}$$

$= \frac{1}{3} \cdot \frac{1}{3} \cdot \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right){\left( {\frac{1}{3}} \right)^2}{\left( {\frac{2}{3}} \right)^2} = \frac{{24}}{{{3^6}}}$

$${\text{P}}(X = 6) = \left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right){\left( {\frac{1}{3}} \right)^4}{\left( {\frac{2}{3}} \right)^2} = \frac{{40}}{{{3^6}}}$$

so required probability is $$\frac{{24}}{{{3^6}}} \cdot \frac{{{3^6}}}{{40}} = \frac{3}{5}$$     A1

[2 marks]

Total [10 marks]

## Examiners report

Realising that this was a problem about the Negative Binomial distribution was the crucial thing to realise in this question. All parts of the syllabus do need to be covered.

## Question

The random variable X has the negative binomial distribution NB(3, p) .

Let $$f(x)$$ denote the probability that X takes the value x .

(i)     Write down an expression for $$f(x)$$ , and show that

$\ln f(x) = 3\ln \left( {\frac{p}{{1 – p}}} \right) + \ln (x – 1) + \ln (x – 2) + x\ln (1 – p) – \ln 2{\text{ .}}$

(ii)     State the domain of f .

(iii)     The domain of f is extended to $$]2,{\text{ }}\infty [$$ . Show that

$$\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x – 1}} + \frac{1}{{x – 2}} + \ln (1 – p){\text{ .}}$$

[7]
a.

Jo has a biased coin which has a probability of 0.35 of showing heads when tossed. She tosses this coin successively and the $${3^{{\text{rd}}}}$$ head occurs on the $${Y^{{\text{th}}}}$$ toss. Use the result in part (a)(iii) to find the most likely value of Y .

[5]
b.

## Markscheme

(i)     $$f(x) = \left( {\begin{array}{*{20}{c}} {x – 1} \\ 2 \end{array}} \right){p^3}{(1 – p)^{x – 3}}$$     M1A1

Note: Award M1A0 for $$f(x) = \left( {\begin{array}{*{20}{c}} {x – 1} \\ 2 \end{array}} \right){p^3}{q^{x – 3}}$$

taking logs,     M1

$$\ln f(x) = \left( {\ln \left( {\begin{array}{*{20}{c}} {x – 1} \\ 2 \end{array}} \right){p^3}(1 – p){}^{x – 3}} \right)$$

$$= \ln \left( {\frac{{(x – 1)(x – 2)}}{2} \times {p^3}{{(1 – p)}^{x – 3}}} \right)$$     A1

Note: Award A1 for simplifying binomial coefficient, seen anywhere.

$$= \ln \left( {\frac{{(x – 1)(x – 2)}}{2} \times {p^3}\frac{{{{(1 – p)}^x}}}{{{{(1 – p)}^3}}}} \right)$$     A1

Note: Award A1 for correctly splitting $${{{(1 – p)}^{x – 3}}}$$ , seen anywhere.

$$= 3\ln \left( {\frac{p}{{1 – p}}} \right) + \ln (x – 1) + \ln (x – 2) + x\ln (1 – p) – \ln 2$$     AG

(ii)     the domain is {3, 4, 5, …}     A1

Note: Do not accept $$x \geqslant 3$$

(iii)     differentiating with respect to x ,     M1

$$\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x – 1}} + \frac{1}{{x – 2}} + \ln (1 – p)$$     AG

[7 marks]

a.

setting $$f'(x) = 0$$ and putting p = 0.35 ,

$$\frac{1}{{x – 1}} + \frac{1}{{x – 2}} + \ln 0.65 = 0$$     M1A1

solving, x = 6.195…     A1

we need to check x = 6 and 7

f (6) = 0.1177… and f (7) = 0.1148…     A1

the most likely value of Y is 6     A1

Note: Award the final A1 for the correct conclusion even if the previous A1 was not awarded.

[5 marks]

b.

## Examiners report

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being $$x \geqslant 3$$ . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.

a.

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being $$x \geqslant 3$$ . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.

b.

## Question

The random variable X has the negative binomial distribution NB(3, p) .

Let $$f(x)$$ denote the probability that X takes the value x .

(i)     Write down an expression for $$f(x)$$ , and show that

$\ln f(x) = 3\ln \left( {\frac{p}{{1 – p}}} \right) + \ln (x – 1) + \ln (x – 2) + x\ln (1 – p) – \ln 2{\text{ .}}$

(ii)     State the domain of f .

(iii)     The domain of f is extended to $$]2,{\text{ }}\infty [$$ . Show that

$$\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x – 1}} + \frac{1}{{x – 2}} + \ln (1 – p){\text{ .}}$$

[7]
a.

Jo has a biased coin which has a probability of 0.35 of showing heads when tossed. She tosses this coin successively and the $${3^{{\text{rd}}}}$$ head occurs on the $${Y^{{\text{th}}}}$$ toss. Use the result in part (a)(iii) to find the most likely value of Y .

[5]
b.

## Markscheme

(i)     $$f(x) = \left( {\begin{array}{*{20}{c}} {x – 1} \\ 2 \end{array}} \right){p^3}{(1 – p)^{x – 3}}$$     M1A1

Note: Award M1A0 for $$f(x) = \left( {\begin{array}{*{20}{c}} {x – 1} \\ 2 \end{array}} \right){p^3}{q^{x – 3}}$$

taking logs,     M1

$$\ln f(x) = \left( {\ln \left( {\begin{array}{*{20}{c}} {x – 1} \\ 2 \end{array}} \right){p^3}(1 – p){}^{x – 3}} \right)$$

$$= \ln \left( {\frac{{(x – 1)(x – 2)}}{2} \times {p^3}{{(1 – p)}^{x – 3}}} \right)$$     A1

Note: Award A1 for simplifying binomial coefficient, seen anywhere.

$$= \ln \left( {\frac{{(x – 1)(x – 2)}}{2} \times {p^3}\frac{{{{(1 – p)}^x}}}{{{{(1 – p)}^3}}}} \right)$$     A1

Note: Award A1 for correctly splitting $${{{(1 – p)}^{x – 3}}}$$ , seen anywhere.

$$= 3\ln \left( {\frac{p}{{1 – p}}} \right) + \ln (x – 1) + \ln (x – 2) + x\ln (1 – p) – \ln 2$$     AG

(ii)     the domain is {3, 4, 5, …}     A1

Note: Do not accept $$x \geqslant 3$$

(iii)     differentiating with respect to x ,     M1

$$\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x – 1}} + \frac{1}{{x – 2}} + \ln (1 – p)$$     AG

[7 marks]

a.

setting $$f'(x) = 0$$ and putting p = 0.35 ,

$$\frac{1}{{x – 1}} + \frac{1}{{x – 2}} + \ln 0.65 = 0$$     M1A1

solving, x = 6.195…     A1

we need to check x = 6 and 7

f (6) = 0.1177… and f (7) = 0.1148…     A1

the most likely value of Y is 6     A1

Note: Award the final A1 for the correct conclusion even if the previous A1 was not awarded.

[5 marks]

b.

## Examiners report

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being $$x \geqslant 3$$ . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.

a.

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being $$x \geqslant 3$$ . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.

b.

## 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).

[1]
a.

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

[1]
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).

[1]
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 $${\text{P}}(Y = y) = \frac{1}{{36}}$$.

[1]
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 $${\text{P}}(Y \leqslant 6)$$ .

[2]
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

When Ben shoots an arrow, he hits the target with probability 0.4. Successive shots are independent.

Find the probability that

(i)     he hits the target exactly 4 times in his first 8 shots;

(ii)     he hits the target for the $${4^{{\text{th}}}}$$ time with his $${8^{{\text{th}}}}$$ shot.

[6]
a.

Ben hits the target for the $${10^{{\text{th}}}}$$ time with his $${X^{{\text{th}}}}$$ shot.

(i)     Determine the expected value of the random variable X.

(ii)     Write down an expression for $${\text{P}}(X = x)$$ and show that

$\frac{{{\text{P}}(X = x)}}{{{\text{P}}(X = x – 1)}} = \frac{{3(x – 1)}}{{5(x – 10)}}.$

(iii)     Hence, or otherwise, find the most likely value of X.

[9]
b.

## Markscheme

(i)     the number of hits, $$X \sim {\text{B(8, 0.4)}}$$     (A1)

$$P(X = 4) = \left( {\begin{array}{*{20}{c}} 8 \\ 4 \end{array}} \right) \times {0.4^4} \times {0.6^4}$$     (M1)

= 0.232     A1

Note: Accept any answer that rounds to 0.23.

(ii)     let the $${4^{{\text{th}}}}$$ hit occur on the $${Y^{{\text{th}}}}$$ shot so that $$Y \sim {\text{NB(4, 0.4)}}$$     (A1)

$$P(Y = 8) = \left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right) \times {0.4^4} \times {0.6^4}$$     (M1)

= 0.116     A1

Note: Accept any answer that rounds to 0.12.

[6 marks]

a.

(i)     $$X \sim {\text{NB(10, 0.4)}}$$     (M1)

$${\text{E}}(X) = \frac{{10}}{{0.4}} = 25$$     A1

(ii)     let $${{\text{P}}_x}$$ denote $${\text{P}}(X = x)$$

$${P_x} = \left( {\begin{array}{*{20}{c}} {x – 1} \\ 9 \end{array}} \right) \times {0.4^{10}} \times {0.6^{x – 10}}$$     A1

$$\frac{{{P_x}}}{{{P_{x – 1}}}} = \frac{{\left( {\begin{array}{*{20}{c}} {x – 1} \\ 9 \end{array}} \right) \times {{0.4}^{10}} \times {{0.6}^{x – 10}}}}{{\left( {\begin{array}{*{20}{c}} {x – 2} \\ 9 \end{array}} \right) \times {{0.4}^{10}} \times {{0.6}^{x – 11}}}}$$     M1A1

$$= \frac{{(x – 1)!}}{{9!(x – 10)!}} \times \frac{{9!(x – 11)! \times 0.6}}{{(x – 2)!}}$$     A1

Note: Award A1 for correct evaluation of combinatorial terms.

$$= \frac{{3(x – 1)}}{{5(x – 10)}}$$     AG

(iii)     $${{\text{P}}_x} > {{\text{P}}_{x – 1}}$$ as long as

$$3x – 3 > 5x – 50$$     (M1)

i.e. $$x < 23.5$$     (A1)

the most likely value is 23     A1

Note: Allow solutions based on creating a table of values of $${{\text{P}}_x}$$.

[9 marks]

b.

## Examiners report

Part (a) was well answered in general although some candidates were unable to distinguish between the binomial and negative binomial distributions.

a.

In (b)(ii), most candidates knew what to do but algebraic errors were not uncommon. Candidates often used equal instead of inequality signs and this was accepted if it led to $$x = 23.5$$. The difficulty for these candidates was whether to choose $$23$$ or $$24$$ for the final answer and some made the wrong choice. Some candidates failed to see the relevance of the result in (b)(ii) to finding the most likely value of $$X$$ and chose an ‘otherwise’ method, usually by creating a table of probabilities and selecting the largest.

b.

## Question

Eric plays a game at a fairground in which he throws darts at a target. Each time he throws a dart, the probability of hitting the target is $$0.2$$. He is allowed to throw as many darts as he likes, but it costs him $$1$$ a throw. If he hits the target a total of three times he wins $$10$$.

Find the probability he has his third success of hitting the target on his sixth throw.

[3]
a.

(i)     Find the expected number of throws required for Eric to hit the target three times.

(ii)     Write down his expected profit or loss if he plays until he wins the $$10$$.

[3]
b.

If he has just $$8$$, find the probability he will lose all his money before he hits the target three times.

[3]
c.

## Markscheme

METHOD 1

let $$X$$ be the number of throws until Eric hits the target three times

$$X \sim {\text{NB(3, 0.2)}}$$     (M1)

$${\text{P}}(X = 6) = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){0.8^3} \times {0.2^3}$$     (A1)

$$= 0.04096\;\;\;\left( { = \frac{{128}}{{3125}}} \right)\;\;\;$$(exact)     A1

METHOD 2

let $$X$$ be the number of hits in five throws

$$X$$ is $${\text{B}}(5,{\text{ }}0.2)$$     (M1)

$${\text{P}}(X = 2) = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){0.2^2} \times {0.8^3}\;\;\;(0.2048)$$     (A1)

$$P$$(3rd hit on 6th throw) $$= \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){0.2^2} \times {0.8^3} \times 0.2 = 0.04096\left( { = \frac{{128}}{{3125}}} \right)\;\;\;$$(exact)     A1

[3 marks]

a.

(i)     $${\text{expected number of throws}} = \frac{3}{{0.2}} = 15$$     (M1)A1

(ii)     $${\text{profit}} = (10 – 15) = – \ 5{\text{ or loss}} = \ 5$$     A1

[3 marks]

b.

METHOD 1

let $$Y$$ be the number of times the target is hit in $$8$$ throws

$$Y \sim {\text{B}}(8,{\text{ }}0.2)$$     (M1)

$${\text{P}}(Y \le 2)$$     (M1)

$$= 0.797$$     A1

METHOD 2

let the $${3^{{\text{rd}}}}$$ hit occur on the $${Y^{{\text{th}}}}$$ throw

$$Y{\text{ is NB}}(3,{\text{ }}0.2)$$     (M1)

$${\text{P}}(Y > 8) = 1 – {\text{P}}(Y \le 8)$$     (M1)

$$= 0.797$$     A1

[3 marks]

Total [9 marks]

c.

## Examiners report

Part (a) was well answered, using the negative binomial distribution $$NB(3,{\text{ }}0.2)$$, by many candidates. Some candidates began by using the binomial distribution $$B(5,{\text{ }}0.2)$$ which is a valid method as long as it is followed by multiplying by 0.2 but this final step was not always carried out successfully.

a.

Part (b) was well answered by the majority of candidates.

b.

In (c), candidates who used the binomial distribution $$B(8,{\text{ }}0.2)$$ were generally successful. Candidates who used the negative binomial distribution $$Y \approx NB(3,{\text{ }}0.2)$$ to evaluate $$P(Y > 8)$$ were usually unsuccessful because of the large amount of computation involved.

c.

## 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 $${\text{P}}(Y = y) = {{\text{e}}^{ – y}}(1 – {{\text{e}}^{ – 1}}),{\text{ }}y \in \mathbb{N}$$.

[4]
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.

[8]
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 $$H(t)$$.

[1]
a.

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

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

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

[10]
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.

[N/A]

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 $$0 \leqslant t \leqslant 2$$.

[3]
a.

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

[2]
b.i.

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

[2]
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

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 $$0 \leqslant t \leqslant 2$$.

[3]
a.

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

[2]
b.i.

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

[2]
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

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 $$0 \leqslant t \leqslant 2$$.

[3]
a.

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

[2]
b.i.

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

[2]
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 $${G’’_X}(t)$$.

[2]
a.i.

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

[3]
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 $$\lambda + \mu$$.

[3]
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 $$n \geqslant x$$.

[5]
c.i.

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

[2]
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.

[N/A]

b.

[N/A]

c.i.

[N/A]

c.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 $${G’’_X}(t)$$.

[2]
a.i.

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

[3]
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 $$\lambda + \mu$$.

[3]
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 $$n \geqslant x$$.

[5]
c.i.

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

[2]
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.

[N/A]

b.

[N/A]

c.i.

[N/A]

c.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 $${G’’_X}(t)$$.

[2]
a.i.

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

[3]
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 $$\lambda + \mu$$.

[3]
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 $$n \geqslant x$$.

[5]
c.i.

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

[2]
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.

[N/A]

b.

[N/A]

c.i.

[N/A]

c.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 $${G’’_X}(t)$$.

[2]
a.i.

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

[3]
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 $$\lambda + \mu$$.

[3]
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 $$n \geqslant x$$.

[5]
c.i.

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

[2]
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.

[N/A]

b.

[N/A]

c.i.

[N/A]

c.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 $${G’’_X}(t)$$.

[2]
a.i.

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

[3]
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 $$\lambda + \mu$$.

[3]
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 $$n \geqslant x$$.

[5]
c.i.

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

[2]
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.

[N/A]

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.

[1]
a.

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

[4]
b.

Find $$G’\left( t \right)$$.

[2]
c.

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

[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.

## 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.

[1]
a.

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

[4]
b.

Find $$G’\left( t \right)$$.

[2]
c.

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

[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.