Notes

(i) B1: Both hypotheses correct in terms of \(\lambda\) (do not accept \(\mu\) or \(\bar{x}\)). May be seen in (a)(ii)/(iii)

(ii) M1: Sight or use of \(X \sim \mathrm{Po}(7)\) [may be implied by a correct probability to 2 s.f. in (a)]

A1ft: Correct CR clearly stated (A0 if given only as a probability)
[ft their hypotheses from (a)(i), 2-tailed CR is \(X \leqslant 2, X \geqslant 13\)]

(iii) B1ft: awrt 0.0818, ft the probability associated with their CR

Notes

M1: Sight or use of \(Y \sim \mathrm{Po}(5)\), condone lambda = 5 as a probability statement
[implied by a correct probability e.g. 0.124…, 0.265…, 0.0404…, 0.0734…]

M1: Attempt to find \(\mathrm{P}(Y \geqslant 4)\) with \(Y \sim \mathrm{Po}(5)\) (ft from their CR)

A1: awrt 0.735 (correct answer implies full marks)

Notes

B1: Stating Geometric or Geo (parameter not required, ignore parameter if given)

Notes

M1: Sight of \(T \sim \mathrm{Geo}(\mathrm{e}^{-\lambda})\)   or   P(zero glitches) = \(\mathrm{e}^{-\lambda}\)   [condone \(\mathrm{P}(T = 0) = \mathrm{e}^{-\lambda}\)]

A1cso*: \(\mathrm{P}(T \leqslant n)\) o.e.   or   stating that Power = 1 – P(Type II error)
leading to \(1 - (1 - \mathrm{e}^{-\lambda})^n\) [working must be fully correct with no errors]

Notes

B1: \((1 - \mathrm{e}^{-2.75})^n \lt 0.2\) (condone use of = or \(\leqslant\) instead of \(\lt\))

M1: Correct process to reach a linear equation involving logs. May be implied by awrt 24.4 or awrt 3.38

A1: Solves equation to obtain awrt 24.4

A1: (\(n =\)) 25 (\(n \geqslant 25\) is A0)
Award full marks for 25 if not from clearly incorrect working

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

(i) M1: selecting NB(4, 0.15) or use of a correct method, may be implied by a correct answer

A1: awrt 0.0364

(ii) M1: selecting B(18, 0.15) or use of a correct method, may be implied by a correct answer

A1: awrt 0.159

Notes

B1: [\(\mathrm{E}(X)\)] \(= \dfrac{4}{p}\)

M1: for an equation or inequality with their expression for mean and 32
allow mean = 32   or   mean \(\gt\) 32
mean \(\leqslant\) 32   or   mean \(\lt\) 32   is M0

A1: 0.125 or exact equivalent
Must not come from clearly incorrect working
NB: \(p \leqslant \dfrac{1}{8}\) is B1M1A0
M0 implies A0 (method mark cannot be implied here)

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 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 \(\mathrm{G}'_X(1)\))

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.

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)

Notes

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

Notes

B1 for both “geometric” or “Geo” and correct parameter of \(\dfrac{1}{6}\). Probability must be seen in (a)

Notes

M1 for a correct method, may be implied by a correct expression

A1cso* for a correct solution leading to printed answer with no incorrect working seen

Notes

1^st M1 for correct use of formula for \(\mathrm{E}(X)\) or \(\mathrm{Var}(X)\)

1^st A1 for both correct must see value of 6 for \(\mathrm{E}(X)\) and at least correct formula used for \(\mathrm{Var}(X)\) the value of 30 may be implied by 2^nd A1

2^nd M1 for use of CLT to get a normal distribution with correct mean and variance \(\neq\) their 30
Condone use of \(X\) (or their letter) instead of \(\overline{X}\) or seeing \(\mathrm{N}\left(6, \sqrt{\dfrac{30}{64}}\right)\)

2^nd A1 for a correct normal distribution stated or used. May be implied by correct answer.

3^rd A1 for awrt 0.681 but allow 0.68 or 0.680 if correct distribution is seen

Notes

B1 for both hypotheses correct in terms of \(p\)

M1 for sight or use of \(\mathrm{P}(X \geqslant 16)\) (may be implied)
or correct expression with logs (use of logs may be implied by 17.43…)

1^st A1 for 0.065 or better but accept 0.06 if a correct expression is seen, or correct CR

2^nd A1 for a correct conclusion mentioning probability or biased
Condone ‘evidence suggests probability is equal to \(\dfrac{1}{6}\)’
NB: Must have correct probability or correct CR to gain final A1.
NB: Send responses comparing 0.935 with 0.95 to review

Notes

(i) M1 for selecting the correct model. Stated or used which may be implied by ans.

A1 for 0.1024 or \(\frac{64}{625}\) (accept 0.102)   (correct answer scores 2 out of 2)

(ii) M1 for selecting the correct model. Stated or used which may be implied by ans.
Allow \(0.2 \times \mathrm{P}(V = 2)\) from \(V \sim \mathrm{B}(7, 0.2)\)

A1 for awrt 0.0551   (correct answer scores 2 out of 2)

(iii) M1 for selecting the correct model. Stated or used may be implied by ans of 0.967(2)

A1 for awrt 0.0881   (correct answer scores 2 out of 2)

Notes

1^st M1 for a correct distribution and prob. expression for \(\mathrm{P}(R)\)   (may be implied by 2^nd M1)

2^nd M1 for a correct numerical expression for \(\mathrm{P}(R)\)   (allow any equivalent expression)   (corrected from the printed mark scheme: the binomial alternative printed \(1 - \mathrm{P}(Y = 0)\); with \(X \sim \mathrm{B}(4, 0.2)\) it is \(1 - \mathrm{P}(X = 0)\))

1^st A1 for awrt 0.590 or \(\frac{369}{625}\) (accept 0.59 or better)   awrt 0.590 implies M1M1A1

3^rd M1 for a correct distribution and prob. expression for \(\mathrm{P}(Y)\)   (may be implied by 4^th M1)

4^th M1 for a correct numerical expression for \(\mathrm{P}(Y)\)   (allow any equivalent expression)

2^nd A1 for awrt 0.580 or (accept 0.58 or better)   awrt 0.580 implies M1M1A1

3^rd A1 dep on all other marks for \(R\) or correct description in words   Condone \(\mathrm{P}(R) \gt \mathrm{P}(Y)\)

Notes

M1 for identifying the NegBin distribution (allow NB for NegBin)

A1 for \(r = 2\) and \(p = \tfrac{1}{3}\)

Notes

B1ft for identifying a suitable definition for \(X\) using a (fair) die, with \(p = \tfrac{1}{3}\) and the second occurrence of the event, only ft their NegBin distribution in (a). A finite number of rolls is B0

Notes

(i) 1^st M1 for attempt to differentiate quotient or product. At least one \(uv'\) style term correct.

1^st A1 for a fully correct first derivative (needn’t be simplified)

2^nd A1 for \(\mathrm{E}(X) = 6\)   NB this A1 depends on M1 only but M1A0A1 is possible

(ii) 2^nd M1 for attempt to diff’ quotient or product again. At least one \(uv'\) style term correct.

3^rd A1 for 42 (may be given for incorrect \(\mathrm{G}''\) provided their \(\mathrm{G}''(1)\) gives 42 and M1 scored)
Note all powers of \((3 - 2t)\) equal 1 when \(t = 1\) is substituted so can be used as a check

3rd M1 for correct use of pgf to find \(\mathrm{Var}(X)\)

4^th A1 dep on M3 for 12

Notes

M1 for writing pgf in suitable form to carry out binomial expansion

1^st A1 for a correct expression for coefficient of \(t^{19}\)

2^nd A1 for \(\frac{32}{243}\) or exact equivalent

Alternative

ALT M1 for identifying connection \(Y = 3X + 4\)
1^st A1 for a correct numerical probability expression for \(\mathrm{P}(X = 5)\)

Notes

1^st M1 for selecting the correct negative binomial model. May be implied by correct mean or variance
NegBin on its own is M0

1^st A1 for mean = 24

2^nd A1 for variance = 120    \(\sigma = 120\) is A0 unless recovered

2^nd M1 for writing or using of normal with mean 24 (may be implied by correct answer)
ft their mean which may come from any distribution

3^rd M1 for writing or using normal with standard deviation \(= \sqrt{\dfrac{120}{32}}\ \left[= \sqrt{3.75}\right]\)
ft \(\dfrac{\text{their } \sigma}{\sqrt{32}}\) where \(\sigma\) may come from any distribution
(may be implied by correct answer)

2^nd A1 for awrt 0.951   (correct answer scores 6 out of 6)

Notes

1^st B1 correct mean

2^nd B1 correct Var   may be implied by sight of \(\frac{7}{135}\) in distribution of \(\overline{X}\)

1^st M1 For use of CLT (must see \(\overline{X}\) and Normal with mean correct ft) or sight of \(\mathrm{N}\left(\dfrac{10}{3}, \dfrac{7}{135}\right)\) or \(\mathrm{N}\left(\text{“}\dfrac{10}{3}\text{”}, \dfrac{\text{“}70\text{”}}{\text{“}9\text{”} \times 150}\right)\) with any letter
Allow 3.33 or better for \(\frac{10}{3}\) and 7.78 or better for \(\frac{70}{9}\)
May be implied by 2^nd M1

2^nd M1 Using the normal distribution to find \(\mathrm{P}(\overline{X} \lt 3.45)\) ft their \(\text{“}\dfrac{10}{3}\text{”}\) and \(\dfrac{\text{“}\dfrac{70}{9}\text{”}}{150}\)
May be implied by correct answer.

A1 awrt 0.696
Correct answer with no incorrect working scores 5/5

Alternative (Use of \(Y = \sum X\))

Notes

(i) M1 Correct method to find \(\mathrm{P}(W = 6)\) eg \((p)^5(1 - p)\) for \(p = 0.11\) or 0.89

A1 awrt 0.0614   (Correct ans with no incorrect working 2/2)

(ii) M1 Correct method to find \(\mathrm{P}(W \leqslant 5)\)

A1 awrt 0.442   (Correct ans with no incorrect working 2/2)

(iii) M1 For using the model \(\mathrm{B}(6, 0.11)\) allow \(\mathrm{B}(6, 0.89)\) [Implied by 0.0017 or awrt 0.114]

A1 awrt 0.00174   (Correct ans with no incorrect working 2/2)

(iv) In part (iv) we can accept correct expressions or values for probabilities

1^st M1 For using a negative binomial model implied by correct \(\mathrm{P}(Y = 5)\) or \(\mathrm{P}(Y = 6)\)

2^nd M1 Correct method to find \(\mathrm{P}(Y \leqslant 6)\)

3^rd M1 At least two correct terms or \(1 - 0.99817..\) from \(1 - \mathrm{P}(V \leqslant 3)\)

A1 awrt 0.00183

Notes

1^st M1 Forming the correct probability of Zac winning or identify \(a\) and \(r\) of GP
Allow for \(p = (0.11) \times 0 + (1 - 0.11)(1 - p)\)

2^nd M1 Using sum to infinity of a GP
Allow for \(p = \dfrac{0.89}{1 + 0.89}\)

A1* Previous method marks must be seen leading to an answer 0.471 (NOT awrt 0.471)

Notes

M1: A correct method to find \(\mathrm{P}(X \geqslant 3)\)

A1: awrt 0.804

Notes

M1: For selecting the appropriate model negative binomial or binomial with an extra trial

A1: awrt 0.0236

Notes

M1: Forming an equation for the mean and variance. At least one correct.

A1: Both equations correct
Allow M1 A1 if both equations correct with the same number subst for \(n\)

M1: Solving the 2 equations leading to \(1 - p = 99p\) oe Allow \(p - p^2 = 99p^2\) ft their 4400 and 660 Allow \(1 - p = 0.15p\)

A1: 0.01

Notes

B1: Both hypotheses correct using correct notation allow eg \(p \gt 0.93\)   (corrected from the printed mark scheme, whose notes label this mark M1; the scheme awards it as B1)

M1: Realising the need to use \(\mathrm{Geo}(0.07)\) ft their Hypotheses

M1: Using the model to find \(\mathrm{P}(J \geqslant c)\)   Condone \((1 - 0.07)^c \lt 0.1\) ft their \(0.07 \neq 0.93\)
ALT \(\mathrm{P}(J \geqslant 32) = 0.1[054\ldots]\) or \(\mathrm{P}(J \geqslant 33) = 0.09[8\ldots]\) Implied by correct CR

M1: For a valid method to solve the inequality or \(\mathrm{P}(J \geqslant 32) = 0.1[054]\) and \(\mathrm{P}(J \geqslant 33) = 0.09[81]\) Implied by correct CR

A1: Correct CR (any letter) A0 if given as a probability statement. Must be integer

Notes

M1: Comparing 34 with their CR eg \(34 \gt 33\)   \(34 \geqslant 33\) or \(\mathrm{P}(J \geqslant 34) = 0.09[12]\)

A1: Fully correct conclusion in context. Allow Jerry’s belief is true. Allow probability for proportion

Notes

M1: Realising they need to find P(their CR in (d))   Allow \(1 - \mathrm{P}(J \leqslant 32)\)

M1: For a Correct method. Allow \(1 - 0.2981\ldots\) May be implied by \(0.7019\ldots\) If the CR is incorrect \((1 - 0.011)^{\text{“CR”} - 1}\) or \(1 - \{1 - (1 - 0.011)^{\text{“CR”} - 1}\}\) must be seen

A1*: Only award if both method marks awarded.

Notes

M1: Selecting geometric distribution with \(p = 0.2\) and attempting required probability.
Allow \((0.8)^n(0.2)\) to imply M1 with \(n = 6\) or \(n = 3\)

A1: awrt 0.0524   Allow exact fraction \(\dfrac{4096}{78125}\)

Notes

M1: \(\mathrm{P}(X \geqslant 6)\) may be implied by \((1 - p)^5\) or \(1 - (p + pq + pq^2 + pq^3 + pq^4)\)

A1: awrt 0.328   Allow exact fraction \(\dfrac{1024}{3125}\)

Notes

B1: Mean = 5

B1: Standard deviation = \(\sqrt{20}\) o.e. or awrt 4.47

Notes

M1: Determining the probability that Suzanne wins with at least three terms seen (may be implied by 2^nd M1)

M1: Recognising need to sum terms of an infinite geometric series with correct \(r = 0.8^2\) (with numerator less than denominator)

A1: \(\dfrac{5}{9}\)   (allow awrt 0.556)

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

M1 for selecting a suitable model i.e. \(\mathrm{B}(40, p)\) where \(p\) is any probability
Written or used, may be implied by a correct ans or \(0.037429\ldots\) from \(\mathrm{P}(X = 3)\)

A1 for awrt 0.0457  (correct answer only 2/2)

Notes

1^st M1 for selecting a suitable model (\(\mathrm{NB}(3, 0.02)\)) May be implied by a correct expression

2^nd M1 for use of model to form a correct expression

SC  \(p \ne 0.02\) Allow prob of the form \(\dbinom{39}{2} p^3 (1 - p)^{37}\) where \(0 \lt p \lt 1\) scores M0M1

A1 for awrt 0.00281  (accept awrt \(2.81 \times 10^{-3}\)) [correct answer with no working scores 3/3]

Notes

B1 for 150