Notes

(i) M1: For sight or use of the correct geometric model

A1: \(\dfrac{4}{27}\) or awrt 0.148 (correct answer implies full marks)

(ii) M1: For selecting the correct geometric model and use of cumulative probability formula
Implied by \(\dfrac{1}{256}\)

M1: Sight or use of \(\mathrm{P}(F \leqslant 8) - \mathrm{P}(F \leqslant 1)\)

A1: \(\dfrac{127}{256}\) or awrt 0.496 (correct answer implies full marks)

(iii) M1: Sight or use of
\(\mathrm{P}[(F = 3) \cup (B = 3)] = \mathrm{P}(F = 3) + \mathrm{P}(B = 3) - \mathrm{P}(F = 3) \times \mathrm{P}(B = 3)\)
or
\(\mathrm{P}[(F = 3) \cup (B = 3)] = 1 - \mathrm{P}(F \neq 3) \times \mathrm{P}(B \neq 3)\)
or
\(\mathrm{P}[(F = 3) \cup (B = 3)] = \mathrm{P}(B = 3) + \mathrm{P}(F = 3) \times \mathrm{P}(B \neq 3)\) (corrected from the printed mark scheme: the notes print this as \(\mathrm{P}(B = 3) - \mathrm{P}(F = 3) \times \mathrm{P}(B \neq 3)\); the scheme above has \(+\))

M1: \(\text{‘}\dfrac{4}{27}\text{’} + \left[\left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right] - \text{‘}\dfrac{4}{27}\text{’} \times \left[\left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right]\)
or
\(1 - \left[1 - \text{‘}\dfrac{4}{27}\text{’}\right] \times \left[1 - \left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right]\)
ft their \(\mathrm{P}(B = 3)\) from (a)(i)

A1: \(\dfrac{55}{216}\) or awrt 0.255 (correct answer implies full marks)

Notes

M1: For attempting to calculate or consider expected values/means of \(X\) and \(Y\)
Implied by sight of 8 and 4.5

B1: for \(\mathrm{E}(X) = 8\)

M1: Using \(\mathrm{E}(B^2) = \mathrm{Var}(B) + [\mathrm{E}(B)]^2\)
or
\(\mathrm{E}(B^2) = \mathrm{G}'_B(1) + \mathrm{G}''_B(1)\) with a correct PGF given for \(B\)
\(\mathrm{E}(B^2) = 15\) is M1

A1: for \(\mathrm{E}(Y) = 7.5\) or equivalent

A1ft: for interpreting the outcome of their calculations in terms of the problem.
Must clearly choose \(X\) or \(Y\) and refer to its expectation/average being higher
Allow reference to expectation as a comparison e.g. \(8 \gt 7.5\)
ft their \(\mathrm{E}(X)\) and \(\mathrm{E}(Y)\) provided at least one of \(\mathrm{E}(X)\) or \(\mathrm{E}(Y)\) correct
dependent on 1^st M1 being awarded
e.g. candidates can score M1B1M0A0A1 or M1B0M1A1A1 or M1B1M1A0A1
NB: Accept expectations given as totals. E.g. \(8n\) and \(7.5n\), but not \(7.5n^2\) etc

Notes

M1: Attempt at \(\mathrm{E}(X)\), at least two terms correct

A1: Full correct expression for \(\mathrm{E}(X)\), may be unsimplified

Notes

M1: Writing or using \(\mathrm{Var}(Y) = b^2\mathrm{Var}(X)\)

A1: For finding \(b = \dfrac{1}{2}\)

B1: Writing using \(\mathrm{E}(Y) = a + b\mathrm{E}(X)\), or writing a correct probability distribution for \(Y\)

M1: Correct expression for \(\mathrm{E}(Y)\) in terms of \(a\) and \(b\) (or their value of \(b\))
and equating their expression to \(\dfrac{5}{16}\)

A1: \(a = \dfrac{4}{9}\) or exact equivalent

M1: Substituting their values of \(a\) and \(b\) into their expression for \(\mathrm{E}(X)\) where \(a, b \neq \dfrac{5}{16}\)

A1: \(\mathrm{E}(X) = -\dfrac{19}{72}\) or exact equivalent

Notes

M1 for selecting a suitable model (\(\mathrm{B}(20, 0.2)\))

A1 for awrt 0.589

Notes

M1 for stating correct negative binomial or \({}^{19}C_3\, p^3(1 - p)^{16} \times p\) for some \(p\)

A1 for awrt 0.0436

Notes

1^st M1 Correct start to problem - sight or use of \(\mathrm{B}(n, 0.2)\). May be implied by \(\mathrm{E}(S) = 0.2n\)

2^nd M1 use of AP formula with \(a = d = 2\), or 2×AP formula with \(a = d = 1\), or equivalent
NB: must be working in another variable, AP formulae cannot be in terms of \(n\)

1^st A1 \(S^2 + S - n\) or equivalent, must be in a form from which expectation can be found

3^rd M1 for use of \(\mathrm{E}(S) = 0.2n\) and \(\mathrm{Var}(S) = 0.16n\) (must be labelled or used as variance) (corrected from the printed mark scheme, which has \(\mathrm{Var}(S^2) = 0.16n\))

2^nd A1 for correct value for \(\mathrm{E}(S^2)\)

(*) 3^rd A1 for a correct solution only, pulling together everything to get given answer

Notes

1^st M1 for using a suitable prob statement or using a listing method to indicate first non-loss \(S\)

2^nd M1 for solving the inequality or using a listing method to reach \(\mathrm{P}(S \geqslant 4)\)
NB: award first two M marks for \(\mathrm{P}(S \geqslant 4)\) provided it does not come from incorrect working

3^rd M1 for attempting \(\mathrm{P}(S \geqslant 4)\) with \(\mathrm{B}(15, 0.2)\)

A1 for awrt 0.352
NB: solutions stemming from finding values of \(n\) gain no marks

Notes

1^st M1 for sight or use of \(\mathrm{negB}(r, 0.2)\) or sight of \(r(r + 1)\)

1^st A1 for \(r(r + 1) - T\), where \(T\) is defined

2^nd M1 for use of \(\mathrm{E}(T) = 5r\) in an expression of “revenue” – \(5r\)

2^nd A1 for \(r^2 - 4r\)

Notes

1^st M1 for an attempt at \(\mathrm{E}(X)\), at least the final 2 products seen or an answer of 2

2^nd M1 for attempting a correct expression for \(\mathrm{E}(X^2)\), at least 3 non-zero products seen

A1 for 4.9 or exact equivalent

Notes

1^st M1 for attempting a correct expression, at least 3 non-zero products seen (implied by 176.9)

2^nd M1 for a correct method, ft their 176.9 and their 8.9 but must be intending \(\mathrm{E}(X^4)\) and \(\mathrm{E}(X^2)\)

A1 for 97.69 or awrt 97.7

Notes

M1 for a correct attempt (at least 3 correct non-zero terms or products and addition)
division by \(k\) \((k \neq 1)\) is M0

A1 for 0.4 o.e. (correct answer only scores 2 out of 2)

Notes

1^st M1 for a correct attempt at \(\mathrm{E}(X^2)\) (at least 3 correct non-zero products and addition)
Missing brackets around –2 and –1 is M0 unless recovered

2^nd dM1 (dep on 1^st M1) for use of 3.9 = their \(\mathrm{E}(X^2) - [\mathrm{E}(X)]^2\)    ft their \(\mathrm{E}(X) = 0.4\)

1^st A1 for \(a = 0.18\) o.e.

2^nd A1 (dep on 1^st M1 only) for \(b = 0.09\) o.e. or their \(b = 0.45 - 2 \times \text{“}a\text{”}\) (provided both \(a\) and \(b\) are probabilities)

Notes

1^st M1 for identifying at least 2 cases e.g. \(X_1 = 3, X_2 \geqslant 1\) counts as 2 cases
(ignore extras including any incorrect pairs identified)
implied by at least two correct products of probs. or correct ft products of probs.

2^nd M1 for a correct numerical expression for the probability    ft their “0.18”

A1 for 0.198 o.e.

Notes

B1 Correct expression for \(\mathrm{E}(X)\)

Notes

1^st M1 Correct attempt at \(\mathrm{E}(X^2)\) using \(\sum x^2\mathrm{P}(X = x)\) at least 3 correct non-zero products
Allow \((-5)^2\) etc

2^nd M1 Realising that \(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\) needs to be used

3^rd M1 Reducing their equation to a 3 term quadratic.   At least 2 terms correct.
Allow e.g. \(0.16b^2 + 0.4b = 23.76\)   Condone missing “=0”

A1 For 11 only (from the correct equation) so \(-13.5\) must be eliminated
Correct answer with no incorrect working seen scores 4/4

Notes

1^st M1 At least 4 values correct for (\(X^2\) and \(2 - 3X\)) or for (\(X^2 - 2\) and \(-3X\)) or \(X^2 + 3X\) or \(X^2 + 3X - 2\) (o.e.)   Allow for solving equation with one sign error

1^st A1ft All correct or correct ft with their \(b\) but must have \(b \gt 5\) (accurate to 1 sf)
Allow solving equation to get awrt \(-3.6\) and awrt 0.56 or \(\frac{-3 \pm \sqrt{17}}{2}\)   (ft their \(b \gt 5\))
If there are omissions but no errors in the lists of values then if 2^nd M1 and 2^nd A1 are scored then the 1^st M1 and 1^st A1 can be given by implication.

2^nd M1 For identifying the correct values of \(X\) required i.e. \(X = -1\) and \(X = 0\)

2^nd A1 0.35
NB It is possible to score M0A0M1A1 here if their table of values is incorrect
Correct answer with no incorrect working seen scores 4/4
(Allow correct use of their \(b \gt 5\))

Notes

M1: For using the given information to find \(\mathrm{E}(N)\)
ALT \(a + b + c = 0.5\) oe

M1: For use of \(\sum n\mathrm{P}(N = n) = \text{“}3.2\text{”}\)   At least 3 terms correct
ALT \(\sum(4n + 2)\mathrm{P}(N = n) = 14.8 \Rightarrow 2a + 1.2 + 0.5 + 3.5 + 18b + 22c = 14.8\)   At least 3 terms correct

M1: Forming an equation in \(b\) and \(c\) using conditional probability

M1: For using \(\sum n^2\mathrm{P}(N = n)\)   Allow with the letters \(b\) and \(c\)

dM1: Dependent on previous method mark. Correct method to find \(\mathrm{Var}(N)\)

A1*: All previous marks must be awarded and 2.76 stated

Notes

M1: Setting up a new model with the correct fees. At least 3 terms correct. Allow 0.5, 0.7, 0.9, 1

M1: Correct method for calculating E(fee) Allow with the letters \(b\) and \(c\)

A1: 81[p]   No units needed. Allow 0.81 if fees are in pounds

Notes

B1: A correct limitation.

Notes

M1: Attempt at \(\mathrm{E}(X)\) with at least 3 correct products seen

M1: Attempt at \(\mathrm{E}(X^2)\) with at least 3 correct products seen

A1: 9 cao

Alternative

M1: Attempt at \(\mathrm{E}(X)\) with at least 3 correct products seen

M1: Attempt at expression for \(\mathrm{E}((X - \mu)^2) = (-5 - 2)^2 \times \tfrac{1}{12} + (-2 - 2)^2 \times \tfrac{1}{6} + (3 - 2)^2 \times \tfrac{1}{4} + (4 - 2)^2 \times \tfrac{1}{2}\) with at least 3 correct terms

A1: 9 cao

Notes

M1: Finding distribution of \(Y\)

M1: \(\mathrm{P}(X = -2) + \mathrm{P}(X = 3)\) or \(\mathrm{P}(Y = 4) + \mathrm{P}(Y = 7)\)

A1: \(\tfrac{5}{12}\)   (condone awrt 0.417)

Notes

M1: Attempt at \(\mathrm{E}(XY)\) with at least 2 correct terms

A1: 13.5

Notes

(i) M1 for selecting the correct model i.e. \(\mathrm{Geo}(p)\) (May be implied by a correct expression)

A1 for \(\frac{8}{81}\)  (\(= 0.098765\ldots\) accept awrt 0.0988)

(ii) M1 for a suitable strategy to use the geometric model to find a correct expression

A1 for \(\frac{211}{243}\)  (\(= 0.868312\ldots\) accept awrt 0.868)

Notes

M1 for a suitable strategy to find \(\mathrm{E}(B^2)\)  [allow \(\mathrm{G}^{\prime\prime}(1) + \mathrm{G}^{\prime}(1)\)]

B1 for use of the correct formulae to find \(\mathrm{E}(B) = 3\) and \(\mathrm{Var}(B) = 6\)  or  \(\mathrm{G}^{\prime\prime}(1) = 12\)

A1 for 15

SC  Formula for \(\mathrm{E}(B^2)\) Allow M1B1A0 for \(\mathrm{E}(B^2) = \dfrac{2 - p}{p^2}\) (o.e.)

Notes

1^st M1 for choosing a suitable geometric model (sight of \(\mathrm{Geo}\left(\frac{2}{3}\right)\) or at least 3 correct probabilities)

2^nd M1 for realising the need for appropriate expected value and using \(\mathrm{E}(\mathrm{g}(X))\) [Need sum and \(\mathrm{f}(x)\)]
NB simply finding \(\mathrm{e}^{\mathrm{E}(X)} = \mathrm{e}^{1.5} =\) awrt 4.48 is M0 and probably no more marks.

3^rd M1 for a suitable strategy to turn the expression into a sum that can be found

1^st A1 for correct use of sum to infinity of geometric series

2^nd A1 for interpreting the outcome of the calculations in terms of a solution to the problem must choose red and see the awrt 19.3 (and allow ft of their \(\mathrm{E}(B^2) \lt 19\))

Notes

B1 for stating or using mean = 3

1^st M1 for using the given model to attempt \(\mathrm{E}(X^2)\) with at least 3 correct products seen

1^st A1 for \(\mathrm{Var}(X) = 2.6\) or \(\sigma = \sqrt{2.6} = 1.6124\ldots\)  (awrt 1.61)

ALT  Use of pgf  (B1 when mean = 3 seen) (M1 when correct \(\mathrm{G}^{\prime\prime}(t)\) seen with attempt at \(\mathrm{G}^{\prime\prime}(1)\))

2^nd M1 for use of CLT – must use \(\overline{X}\) and normal or sight of \(\mathrm{N}\left(\text{“}3\text{”}, \sqrt{\dfrac{\text{“}2.6\text{”}}{80}}^{\,2}\right)\) with any letter

2^nd A1ft for a correct mean and variance, ft their 3 and their 2.6
This M1A1ft may be implied by sight of correct st. dev. used in a standardisation leading to \(\mathrm{P}(Z \gt 1.39)\) Must see correct use of \(Z\)
NB \(\dfrac{2.6}{80} = 0.0325\) and \(\sqrt{\dfrac{2.6}{80}} = 0.18027\ldots\) so allow e.g. \(\mathrm{N}(3, \text{awrt } (0.180)^2)\)

3^rd A1 for using the normal model to find probability awrt 0.0828

ALT  Use of \(\sum X\)  (If see clear attempt at \(\mathrm{P}(\Sigma X \gt 260)\) condone \(\mathrm{P}(\Sigma X \gt 260.5)\)) then:

2^nd M1 for \(\Sigma X \sim \mathrm{N}(\ldots)\) or any letter \(\sim \mathrm{N}\left(\text{“}240\text{”}, \sqrt{\text{“}2.6\text{”} \times 80}^{\,2}\right)\)

2^nd A1ft for mean \(= \text{“}3\text{”} \times 80 = 240\) and variance \(= \text{“}2.6\text{”} \times 80 = 208\)

May see \(\mathrm{P}(\Sigma X \gt 260.5) = 0.077597\ldots\) but it will only score 2^nd M1 2^nd A1ft and 3^rd A0

Notes

M1 Using \(\dfrac{a\mathrm{E}(X_1) + b\mathrm{E}(X_2)}{n}\) and subst \(\mathrm{E}(X_1) = np\) and \(\mathrm{E}(X_2) = np\)

Acso* Answer given. Need \(p(a + b) = p\) and statement \(a + b = 1\) and no errors

Notes

M1 Using \(\dfrac{a^2\mathrm{Var}(X_1) + b^2\mathrm{Var}(X_2)}{n^2}\) and subst \(\mathrm{Var}(X_1) = np(1 - p)\) – may be implied by \(\dfrac{1}{n^2}\left(a^2np(1 - p) + b^2np(1 - p)\right)\)

A1 correct answer in any form

M1d dep on 1^st M1 Subst \(b = 1 - a\)

A1cso* method must be shown and no errors.

Notes

M1 \(\dfrac{\mathrm{d}}{\mathrm{d}a}(\mathrm{Var})\) (must differentiate with respect \(a\)) or attempt to complete the square

A1 correct diff \(= 0\) or \(2\left(a - \tfrac{1}{2}\right)^2 + \tfrac{1}{2}\)

A1 \(a = 0.5\)

A1 ft for \(b = 1 - a\)

B1 for a reason why minimum

Notes

(i) M1 multiplying out and using \(\mathrm{E}(aX) = a\,\mathrm{E}(X)\) [may use their values of \(a\) and \(b\)]

M1d dependent on previous M being awarded Using \(\mathrm{E}(X^2) = \mathrm{Var}(X) + [\mathrm{E}(X)]^2\)

A1 \(\dfrac{(2a^2 - 2a + 1)p(1 - p)}{n} + p^2\) or \(\dfrac{(a^2 + b^2)p(1 - p)}{n} + p^2\) must be of the form \(p^2 +\) a single term

A1 for a reason why it is not equal \(p^2\) plus statement to say biased.

(ii) B1 Follow on from their expression \(p^2 + \ldots\) with \(a\) and \(b\).

Notes

M1 multiplying out correctly and subst \(np\) for \(\mathrm{E}(X)\) or using \(\mathrm{E}(\hat{p})^2 = \mathrm{Var}(\hat{p}) + \left[\mathrm{E}(\hat{p})\right]^2\)
Allow \(= \dfrac{(2a^2 - 2a + 1)p(1 - p)}{n} + p^2\)

A1 \(np^2(n - 1)\)

A1 \(\dfrac{X_1(X_1 - 1)}{n(n - 1)}\)

NB \(\dfrac{X_1(X_1 - 1)}{n(n - 1)}\) gains all 3 marks.

Notes

(i) M1 either \(\dfrac{1}{n}\mathrm{E}(X)\) or \(\dfrac{1}{n} \times np\)
A1 cso

(ii) M1 either \(\dfrac{1}{n^2}\mathrm{Var}(X)\) or \(\dfrac{1}{n^2} \times np(1-p)\)
A1 cso

Notes

(i) M1 For either \(3a\,\mathrm{E}(\hat{p}_1) + 2a\,\mathrm{E}(\hat{p}_2)\) or \(3ap + 2ap\)
M1 Putting their \(\mathrm{E}(\hat{p}_3) = p\)

(ii) M1 for \(\dfrac{9}{25}\mathrm{Var}(\hat{p}_1) + \dfrac{4}{25}\mathrm{Var}(\hat{p}_2)\)
M1d for substituting (aii) for \(\mathrm{Var}(\hat{p}_1)\) and (aii) with \(m\) instead of \(n\) for \(\mathrm{Var}(\hat{p}_2)\)
A1 cso

Notes

M1 Putting \(\mathrm{Var}(\hat{p}_3) \lt\) their \(\mathrm{Var}(\hat{p}_1)\) leading to an inequality of the form \(\dfrac{n}{m} \lt a\) or \(\dfrac{n}{m} \gt a\) where a is a constant.
M1 Putting \(\mathrm{Var}(\hat{p}_3) \lt\) their \(\mathrm{Var}(\hat{p}_2)\) leading to an inequality of the form \(\dfrac{n}{m} \gt a\) or \(\dfrac{n}{m} \lt a\) where a is a constant.

Notes

1/3 is not in their range in part(c)
M1 attempt to find all 3 variances or eliminating \(\mathrm{Var}(\hat{p}_3)\) with reason and finding the other 2 variances.
A1ft correct estimator chosen.
A1ft correct supporting reason from correct working for their var formulae

SC if 1/3 is in their range in part(c) they may get
B1 for stating \(\hat{p}_3\)
B1dependent on the previous B being awarded- stating smallest variance
award first two marks on epen.