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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]
▶️Answer/Explanation
Markscheme
X is geometric (or negative binomial) A1
[1 mark]
\(G\left( t \right) = \frac{1}{4}t + \frac{1}{4}\left( {\frac{3}{4}} \right){t^2} + \frac{1}{4}{\left( {\frac{3}{4}} \right)^2}{t^3} + \ldots \) 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]
attempt to use product or quotient rule M1
\(G’\left( t \right) = \frac{4}{{{{\left( {4 – 3t} \right)}^2}}}\) A1
[2 marks]
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]
Examiners report
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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]
▶️Answer/Explanation
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]
(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]
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 \(Y\).
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.
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]
▶️Answer/Explanation
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]
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]
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]
Examiners report
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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 \(\mu \).
a.i.Find expressions for \({G’_X}(t)\) and \({G’’_X}(t)\).[2]
a.ii.Hence show that \({\text{Var}}(X) = \lambda \).[3]
b.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]
c.i.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.ii.Identify the probability distribution given in part (c)(i) and state its parameters.[2]
▶️Answer/Explanation
Markscheme
\({G’_X}(t) = \lambda {{\text{e}}^{\lambda (t – 1)}}\) A1
\({G’’_X}(t) = {\lambda ^2}{{\text{e}}^{\lambda (t – 1)}}\) A1
[2 marks]
\({\text{Var}}(X) = {G”_X}(1) + {G’_X}(1) – {\left( {{{G’}_X}(1)} \right)^2}\) (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]
\({G_{X + Y}}(t) = {{\text{e}}^{\lambda (t – 1)}} \times {{\text{e}}^{\mu (t – 1)}}\) 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]
\({\text{P}}(X = x|X + Y = n) = \frac{{{\text{P}}(X = x \cap Y = n – x)}}{{{\text{P}}(X + Y = n)}}\) (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]
\({\text{B}}\left( {n,{\text{ }}\frac{\lambda }{{\lambda + \mu }}} \right)\) A1A1
Note: Award A1 for stating binomial and A1 for stating correct parameters.
[2 marks]
Examiners report
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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]
▶️Answer/Explanation
Markscheme
X is geometric (or negative binomial) A1
[1 mark]
\(G\left( t \right) = \frac{1}{4}t + \frac{1}{4}\left( {\frac{3}{4}} \right){t^2} + \frac{1}{4}{\left( {\frac{3}{4}} \right)^2}{t^3} + \ldots \) 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]
attempt to use product or quotient rule M1
\(G’\left( t \right) = \frac{4}{{{{\left( {4 – 3t} \right)}^2}}}\) A1
[2 marks]
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]
Examiners report
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