A2 June 2019 Q6

EdexcelCurrent spec12 marksProbability Generating Functions

6. The discrete random variable \(X\) has probability generating function

\[\mathrm{G}_X(t) = k \ln\left(\frac{2}{2 - t}\right)\]

where \(k\) is a constant.

(a) Find the exact value of \(k\) (1)

(b) Find the exact value of \(\mathrm{Var}(X)\) (7)

Scheme	Marks	AO
\(\mathrm{G}(1) = 1 \Rightarrow k \ln 2 = 1\) so \(k = \underline{\underline{\dfrac{1}{\ln 2}}}\)	B1	2.1
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(\left\{\mathrm{G}(t) = \dfrac{1}{\ln 2}\left[\ln 2 - \ln(2 - t)\right]\right\} \Rightarrow \mathrm{G}^{\prime}(t) = \dfrac{1}{\ln 2}\left[\dfrac{1}{2 - t}\right]\) or \(\dfrac{1}{\ln 2}(2 - t)^{-1}\)	M1 A1	2.1 1.1b
\([\mathrm{E}(X) =]\ \mathrm{G}^{\prime}(1) = \dfrac{1}{\ln 2}\)	A1	1.1b
\(\mathrm{G}^{\prime\prime}(t) = \dfrac{1}{\ln 2} \times \left[\dfrac{1}{(2 - t)^2}\right]\)	M1 A1	2.1 1.1b
\(\mathrm{Var}(X) = \mathrm{G}^{\prime\prime}(1) + \mathrm{G}^{\prime}(1) - \left[\mathrm{G}^{\prime}(1)\right]^2 = \dfrac{1}{\ln 2} + \dfrac{1}{\ln 2} - \left(\dfrac{1}{\ln 2}\right)^2\)	M1	2.1
\(= \underline{\dfrac{1}{\ln 2}\left(2 - \dfrac{1}{\ln 2}\right)}\)	A1	1.1b
	(7)

Notes

1^st M1 for an attempt to differentiate \(\mathrm{G}(t)\) e.g. \(A(2 - t)^{-1}\) (o.e.)

1^st A1 for a correct first derivative (condone \(k\) or use of \(\frac{1}{\ln 2} =\) awrt 1.44)

2^nd A1 for correct \(\mathrm{E}(X)\) or \(\mathrm{G}^{\prime}(1)\) (allow awrt 1.44 calc: \(1.442695\ldots\) but not \(k\)) seen anywhere

2^nd M1 for attempting second derivative (ft their \(\mathrm{G}^{\prime}(t)\))

3^rd A1 for a correct 2^nd derivative (condone \(k\) or use of \(\frac{1}{\ln 2} =\) awrt 1.44)

3^rd M1 for a correct method for \(\mathrm{Var}(X)\) (some substitution into the correct formula)

4^th A1 for \(\dfrac{1}{\ln 2}\left(2 - \dfrac{1}{\ln 2}\right)\) o.e. but must simplify i.e. collect like terms
[Mark final answer – penalise incorrect log work etc]
NB \(0.8040211\ldots\) is A0 unless exact answer seen

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{P}(X = 3) =\) coefficient of \(t^3\) by Maclaurin need \(\mathrm{G}^{\prime\prime\prime}(0)\)	M1	3.1a
\(\mathrm{G}^{\prime\prime\prime}(t) = \dfrac{1}{\ln 2}\dfrac{2}{(2 - t)^3}\)	A1ft	1.1b
\(\mathrm{P}(X = 3) = \dfrac{\mathrm{G}^{\prime\prime\prime}(0)}{3!}\)	M1	3.2a
\(= \dfrac{\frac{1}{4 \ln 2}}{6} = \dfrac{1}{24 \ln 2} = 0.0601122\ldots\) awrt 0.0601	A1	1.1b
	(4)
	(12 marks)

Notes

1^st M1 for a suitable strategy to solve the problem (finding link with Maclaurin)
Need mention of coefficient of \(t^3\) and [\(\mathrm{G}^{\prime\prime\prime}(t)\) or \(\mathrm{G}^{\prime\prime\prime}(0)\)] (condone \(\mathrm{G}^{\prime\prime\prime}(1)\))

1^st A1ft for 3^rd derivative, ft their 2^nd derivative in (b) (provided \(\mathrm{G}^{\prime\prime}(t)\) not const)
Correct \(\mathrm{G}^{\prime\prime\prime}(t)\) or \(\mathrm{G}^{\prime\prime\prime}(0)\) scores 1^st M1 1^st A1ft

2^nd M1 for translating Maclaurin to probability (a correct expression)

2^nd A1 for \(\frac{1}{24 \ln 2}\) or awrt 0.0601

ALT (c) Log series

1^st M1 attempt to write \(\mathrm{G}(t)\) in suitable form as far as: \(k\left[\ln 2 - \ln\left(2\left[1 - \tfrac{t}{2}\right]\right)\right]\)

1^st A1 reaching \(-k \ln\left(1 - \tfrac{t}{2}\right)\)

2^nd M1 use of \(-\ln(1 - x)\) series (some correct substitution) NB \(\mathrm{G}(t) = \tfrac{1}{\ln 2}\left(\tfrac{t}{2} + \tfrac{t^2}{8} + \tfrac{t^3}{24} + \ldots\right)\)