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

## 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 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 $$X$$.

a.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]

b.(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]

## 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]

## 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 $$J(t)$$.

a.Write down an expression for $$J(t)$$ in terms of $$G(t)$$ and $$H(t)$$.[1]

b.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]

## 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]

## Question

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

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

a.Find the cumulative distribution function $$F(t)$$, for $$0 \leqslant t \leqslant 2$$.[3]

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

b.ii.Given that $$P(T < a) = 0.75$$, find the value of $$a$$.[2]

## 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]

## Question

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

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

a.Find the cumulative distribution function $$F(t)$$, for $$0 \leqslant t \leqslant 2$$.[3]

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

b.ii.Given that $$P(T < a) = 0.75$$, find the value of $$a$$.[2]

## 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]

## Question

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

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

a.Find the cumulative distribution function $$F(t)$$, for $$0 \leqslant t \leqslant 2$$.[3]

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

b.ii.Given that $$P(T < a) = 0.75$$, find the value of $$a$$.[2]

## 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

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

a.Find the cumulative distribution function $$F(t)$$, for $$0 \leqslant t \leqslant 2$$.[3]

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

b.ii.Given that $$P(T < a) = 0.75$$, find the value of $$a$$.[2]

## 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

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.

a.State the distribution of X.[1]

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

c.Find $$G’\left( t \right)$$.[2]

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

## 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]

## 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.

a.State the distribution of X.[1]

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

c.Find $$G’\left( t \right)$$.[2]

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

## 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]

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