A2 June 2024 Q6

EdexcelCurrent spec16 marksGeometric & Negative Binomial Probability Generating Functions

6. The random variable \(X\) has probability generating function \(\mathrm{G}_X(t)\) where

\[\mathrm{G}_X(t) = \frac{1}{\sqrt{4 - 3t}}\]

(a) Use calculus to find \(\mathrm{Var}(X)\)
Show your working clearly. (6)

(b) Find the exact value of \(\mathrm{P}(X \leqslant 2)\) (4)

The independent random variables \(X_1\) and \(X_2\) each have the same distribution as \(X\)

The random variable \(Y = X_1 + X_2 + 1\)

(d) Hence, or otherwise, find \(\mathrm{P}(X_1 + X_2 \gt 5)\) (2)

Mark scheme (a)

Scheme	Marks	AO
[\(\mathrm{G}_X(t) = (4 - 3t)^{-\frac{1}{2}} \Rightarrow\)] \(\mathrm{G}'_X(t) = \dfrac{3}{2}(4 - 3t)^{-\frac{3}{2}}\); [So \(\mathrm{E}(X) =\)] \(\mathrm{G}'_X(1) = \dfrac{3}{2}\)	M1; A1	2.1; 1.1b
\(\mathrm{G}''_X(t) = \dfrac{27}{4}(4 - 3t)^{-\frac{5}{2}}\); so \(\mathrm{G}''_X(1) = \dfrac{27}{4}\)	M1; A1ft	2.1 1.1b
\(\mathrm{Var}(X) = \text{“}\dfrac{27}{4}\text{”} + \text{“}\dfrac{3}{2}\text{”} - \left(\text{“}\dfrac{3}{2}\text{”}\right)^2; \ = \underline{\mathbf{6}}\)	M1; A1	1.1b 1.1b
	(6)

Notes

1^st M1 for attempt to differentiate leading to \(k(4 - 3t)^{-1.5}\); 1^st A1 for \(\mathrm{E}(X) = 1.5\) or exact equivalent

2^nd M1 for attempting to differentiate again leading to \(m(4 - 3t)^{-2.5}\)

2^nd A1ft for \(\dfrac{27}{4}\) or correct ft from their \(k\) provided both Ms are scored

3^rd M1 for a correct method for finding \(\mathrm{Var}(X)\); can ft their \(\dfrac{3}{2}\) and their \(\dfrac{27}{4}\)

3^rd A1 for 6

(corrected from the printed mark scheme: the scheme prints “[So \(\mathrm{E}(X)\) =] \(\mathrm{G}'_X(t) = \dfrac{3}{2}\)”; it should be \(\mathrm{G}'_X(1)\))

Mark scheme (b)

Scheme	Marks	AO
Using Maclaurin: \(\dfrac{\mathrm{G}''_X(0)}{2!} = \left\{\dfrac{1}{2}\right\} \times \text{“}\dfrac{27}{4}\text{”} \times \dfrac{1}{32}\) Using Binomial: [\(\mathrm{G}_X(t) =\)] \(\dfrac{1}{2}\left(1 - \dfrac{3}{4}t\right)^{-\frac{1}{2}}\)	M1	2.1
Maclaurin: [\(\mathrm{P}(X = 2) =\)] \(\dfrac{1}{2} \times \dfrac{27}{4} \times \dfrac{1}{32}\ \left[= \dfrac{27}{256}\right]\) Binomial: \(\dfrac{1}{2}\left(\ldots + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\dfrac{3}{4}t\right)^2 + \ldots\right) = \left[\ldots + \dfrac{27}{256}t^2 + \ldots\right]\)	A1	1.1b
Maclaurin: [\(\mathrm{P}(X = 0) + \mathrm{P}(X = 1) =\)] \(\text{“}\dfrac{3}{2}\text{”} \times \dfrac{1}{8} + \dfrac{1}{2}\) Binomial: \(\dfrac{1}{2}\left(1 + \dfrac{3}{8}t + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\dfrac{3}{4}t\right)^2 + \ldots\right)\)	M1	2.1
\(\mathrm{P}(X \leqslant 2) = \dfrac{203}{256}\)	A1	1.1b
	(4)

Notes

1^st M1 for Maclaurin to find \(\mathrm{P}(X = 2)\) condone \(\text{“}\dfrac{27}{4}\text{”} \times \dfrac{1}{32}\left[= \dfrac{27}{128} \text{ or } 0.2109375\right]\)
or putting in the form \(a(1 - 0.75t)^{-0.5}\)

1^st A1 for a correct unsimplified prob for \(\mathrm{P}(X = 2)\) (may be in binomial expansion) Allow 0.105(468…)

2^nd M1 for use of pgf to find \(\mathrm{P}(X = 1)\) and \(\mathrm{P}(X = 0)\) or attempt 1^st 3 terms of bin expansion

2^nd A1 for \(\dfrac{203}{256}\) or exact equivalent.

Mark scheme (c)

Scheme	Marks	AO
\(\left[\mathrm{G}_W(t) = \mathrm{G}_{X_1}(t) \times \mathrm{G}_{X_2}(t) \times t =\right] \dfrac{1}{\sqrt{4 - 3t}} \times \dfrac{1}{\sqrt{4 - 3t}} \times t\); \(\mathrm{G}_Y(t) = \dfrac{t}{4 - 3t}\)	M1; A1	3.1a/2.1 1.1b
\(\mathrm{G}_Y(t) = \dfrac{\frac{1}{4}t}{1 - \frac{3}{4}t}\); \(\boldsymbol{Y \sim}\) Geo\(\left(\dfrac{1}{4}\right)\)	M1 A1	2.1 3.2a/2.2a
	(4)

Notes

1^st M1 for using product of pgf or multiplication by \(t\)

1^st A1 for correct unsimplified form of pgf

2^nd dM1 (dep. on 1^st M1) for attempting to convert pgf to form given in the formula book
or for stating Geometric alongside a correct PGF for \(Y\) (may be unsimplified)

2^nd A1 for correctly deducing the distribution of \(Y\) as geometric with \(p = 0.25\) [may be seen in (d)]
NB: Final A1 dependent on previous marks being awarded.

(corrected from the printed mark scheme: the scheme also prints \(\mathrm{G}_Y(t) = \dfrac{\frac{1}{4}}{1 - \frac{3}{4}t}\) as an alternative form; this is missing the factor \(t\) and is not the pgf of \(Y\), so it has been left out)

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{P}(Y \gt 6) = \left(1 - \dfrac{1}{4}\right)^6; \quad = \dfrac{729}{4096} = 0.177978\ldots\) awrt 0.178	M1 A1	3.4,1.1b
	(2)
	(16 marks)

Notes

M1 for attempting to use geometric formula or their pgf (using correct coefficients); A1 for awrt 0.178