Notes

M1: for a suitable expression (in their \(p\)) for the probability both are yellow.

A1: for correct probability of yellow from B. Can be implied by values on tree diagram.

Notes

B1: for a correct 1^st column of tree diagram (probabilities from A)

B1ft: (dep on M1 in (i)) for correct 2^nd column of tree diagram (ft their \(\frac{9}{16}\) and \(1 - \frac{9}{16}\))

Notes

M1: for selection of all correct cases (including the \(1 -\) if required). Don’t need P(… may be implied by correct probability products seen (ft their tree diagram)

A1ft: for a correct complete probability expression (ft their tree diagram) or awrt 0.708
Must see product pairs (for ft) or implied by correct values e.g. \(\frac{15}{64} + \frac{7}{48} + \frac{3}{16} + \frac{9}{64}\)
May see \(1 - \mathrm{P}\left(\bar{Y}\,\bar{Y}\right)\) (M1) and \(1 - \dfrac{2}{3} \times \dfrac{\text{``}7\text{''}}{\text{``}16\text{''}}\) (A1ft)

A1: for an exact answer …fraction needn’t be simplified but decimal must have \(\ldots\dot{3}\)

Acc: If scored 2^nd A0 in (b) for 0.708 (or better) then allow A1 in (c) for awrt 0.199

Notes

M1: for a correct ratio of probabilities expression (in symbols) or ft their values seen

A1: for an exact answer NB an answer of awrt 0.199 (0.198529411…) implies M1

Acc: If scored 2^nd A0 in (b) for 0.708 (or better) then allow A1 in (c) for awrt 0.199

Notes

B1 for a correct pair. If more than one pair is given then all must be correct.
\(\mathrm{P}(A)\) and \(\mathrm{P}(C)\) etc is B0 \(\mathrm{P}(A \cap C) = 0\) is B0 but condone things like \(A \cap C = \varnothing\)

Notes

In parts (b) – (d) we will condone poor notation and mark equations/expressions

M1 for all relevant labelled probabilities listed or a correct equation/expression for \(p\)

A1 for \(p = 0.13\)

Notes

In parts (b) – (d) we will condone poor notation and mark equations/expressions

In parts (c) and (d) they can use letter \(p\) or we ft their value for \(p\) provided a probability

1^st M1 for a correct method for \(\mathrm{P}(A \mid B^{\prime})\) in \(q\) (and \(r\)) ft their \(p\). May be done in stages
e.g. find correct expression for \(\mathrm{P}(B^{\prime})\), simplify incorrectly then use \(q\) over this

2^nd M1 for a correct equation for \(q + r\) (o.e.)(ft their \(p\)) Can accept \(r = 0\) and \(q = 0.22\)
NB sight of \(\tfrac{0.22}{0.55}\) will score M1M1

A1 for 0.4 i.e. deducing the maximum value of \(\mathrm{P}(A \mid B^{\prime})\). Allow \(\leqslant 0.4\) or \(\mathrm{P}(A \mid B^{\prime}) = 0.4\)
Can award 3/3 for \(\mathrm{P}(A \mid B^{\prime}) = 0.4\) but not 0.4 alone as it can come from e.g \(\mathrm{P}(C^{\prime})\)

Notes

In parts (b) – (d) we will condone poor notation and mark equations/expressions

In parts (c) and (d) they can use letter \(p\) or we ft their value for \(p\) provided a probability

M1 for a correct equation for \(r\) (or \(q\)) only can have \(p\) or ft their value for \(p\).
May be in stages
e.g. find \(\mathrm{P}(A^{\prime}) = 0.27 + 0.25 + 0.08 + p + r\) but make a slip in getting 0.6 then use this.

1^st A1 for \(r = 0.07\) or \(q = 0.15\)

2^nd A1ft for \(r = 0.07\) and \(q = 0.15\) or values giving \(q + r = 0.22\) provided both \(q\) and \(r\) are probabilities. Obviously, 2^nd A1ft is dependent on the M1

Notes

B1 for 0.33

Notes

B1 for any correct expression. Do not condone P(…

Notes

B1 for awrt 0.0912 (from calculator)

Notes

B1 for a suitable reason mentioning or implying \(H\) and \(T\) are independent
Allow: e.g. “they”/ “each event”/ “\(\mathrm{P}(H)\) and \(\mathrm{P}(T)\)”/ “the variables” and “independent”
B0 for “the results” / “the values” are independent.
Ignore other comments that are not incorrect or contradictory.

Notes

1^st M1 for using model for \(T\) to attempt to find \(\mathrm{P}(T \lt 300)\) e.g. sight of 0.124 or better or sight of \(\pm\left(\dfrac{300 - 330}{26}\right)\) or \(\pm\left(\dfrac{5 - 5.5}{0.433\ldots}\right)\) or \(Z = \pm\,1.15(3\ldots)\)

2^nd M1 for multiplying their two probabilities together ft part (a) and their \(\mathrm{P}(T \lt 300)\) provided both values are probabilities. NB M0M1 is possible here

A1 for awrt 0.0113   [Correct answer with no incorrect working 3/3]

Notes

1^st M1 for standardising 16.3 and setting equal to \(z\) value where \(0.5 \lt |z| \lt 0.6\)

2^nd M1 for standardising 29 and setting equal to \(z\) value where \(1 \lt |z| \lt 1.5\)

3^rd M1 dep on 1^st or 2^nd M1 for solving their two linear eq’ns – reach an eq’n in one variable
May be implied by sight of \(\sigma = 7\) (or better) or \(\mu = 20\) (or better)

For 1^st A mark we must also see one of \(-0.5244\) or 1.2816 (or better) used in their equ’ns OR both \(z\) values correct to 3dp i.e. \(-0.524\) and 1.282

1^st A1 for \(\sigma =\) awrt 7.03 (but see 3^rd case below)

2^nd A1 for \(\mu =\) in [19.95, 20.0] (i.e shouldn’t see something rounding down to 20.0)
Allow 20 from equations with suitable \(z\) values (see examples below)

NB Use of \(-0.524\) and 1.28 [would give 7.0399… and 19.988…] and scores M3A0A1
Use of \(-0.524\) and 1.2816 [would give 7.033… and 19.99…] and scores M3A1A1
Use of \(-0.5244\) and 1.28 [would give 7.038… and 19.99 …] and scores M3A1A1

Both \(z\) values correct to 3dp i.e. \(-0.524\) and 1.282 [should give 7.032 and 19.984] scores A1A1

Typical solution

0.12

Typical solution

Typical solution

Typical solution

Typical solution

Hence events \(Y\) and \(Z\) are not independent.

Typical solution

(i)

0.39

(ii)

(iii)

Typical solution

Hence \(G\) and \(H\) are not independent

Typical solution

The bottom-left Venn diagram: the shaded region is the part of B that also lies in A or in C

Typical solution

P(both bronze) = \(0.2 \times 0.2 = 0.04\)

P(both silver) = \(0.8 \times 0.8 = 0.64\)

P(both same type) = 0.68

Typical solution

P(at least one bronze)

P(both bronze | at least one bronze)

Typical solution

Typical solution

Typical solution

(i)

(ii)

(iii)

Typical solution

Hence events \(C\) and \(T\) are not independent

Notes

M1: May be implied by first A1 (which is implied by use of 4 correct probabilities).

A1: Soi

M1: FT their probabilities, dep \(\Sigma p = 1\) Allow M1 if \(\times 2\) omitted or \(2 \times\) an additional term.
Allow this mark for \(44p^3\) May see e.g.

Leading to an expression of the form given (3 or 4 terms each comprised of 3 of their probabilities multiplied together).

A1: Correct value from correct working implies all 4 marks (i.e. if candidates write \(44p^3\) and \(p = \frac{1}{15} \Rightarrow \mathrm{P}(\ldots) = \frac{44}{3375}\) then score 4/4)
Accept awrt 0.0130 (3sf)

Notes

(a)(i)

B1: awrt 0.556 (3sf)

(a)(ii)

M1: Allow \(1 - \frac{4}{9}\) or \(\frac{5}{9} - \cdots\) (where \(\cdots\) is any other correct work)

M1: May be included in an expression of the form \(1 - \cdots\).

M1: May be included in an expression of the form \(1 - \cdots\).

A1: Some indication of summation must be seen
Or \(1 - \left(\frac{4}{9} + 6 \times \frac{1}{9} \times \frac{1}{8} + 4 \times \frac{1}{9} \times \frac{1}{8} \times \frac{1}{7}\right) = \frac{5}{9} - \frac{6}{72}\left(\frac{1}{12}\right) - \frac{4}{504}\left(\frac{1}{126}\right)\)

Alternative method 1 (counting possibilities for first three digits)

A1: AG

Alternative method (using permutations)

A1: AG

Notes

M1: M1 for \(\frac{13}{28} \div\) their answer to (i) soi. Allow 0.464 for \(\frac{13}{28}\)

A1FT: FT (i) (but A0 for answers outside \([0,1]\))

Notes

M1*: Forming this equation in \(p\), must be fully correct with \(= 1\) soi

M1 dep*: Rearrange their equation to solvable form \(ap^2 + bp + c = 0\) and attempt to solve (may be implied by one or both correct roots)

B1: For sight of \(p = -\frac{3}{2}\) oe provided this root not used in subsequent working. Condone “the other root is negative” or “\(p \gt 0\)”

A1: These values are likely to be seen in, and may be implied by, subsequent working. Note that a correct numerical denominator or final answer also implies this mark.

M1: Either numerator or denominator attempted (correct form with 2 terms in the numerator or 6 terms in the denominator). FT their probabilities.
May be implied by any of:
• a correct expression for one of the numerator or denominator either in \(p\) or with their probabilities
• a correct final answer

M1: Division attempted with a 2-term numerator and 6-term denominator soi either in \(p\) or with their probabilities (may see numerator and denominator computed separately and then an attempt to divide)

A1: oe, need not be simplified

Notes

M1: Must be seen. Condone a correct statement e.g. “The probability of obtaining 2 is not the same as the probability of getting 3” – must refer to specific scores, need not compute probabilities (but if given these must be correct).

A1: A generalised conclusion must be present:

• Allow “probabilities are not the same”
• May come at the beginning e.g. “the chance of each student being chosen is not equal because…”

Alternative using combinations

M1: Must be seen. Must refer to specific scores but need not compute the number of combinations/ways (but if given these must be correct).

A1: A generalised conclusion must be present:

• Accept “student __ is more likely to be chosen than student __, therefore it is not random”
• Accept “some scores are more likely than others”

Notes

(b)(i) B1: Any valid correction to the method, e.g.:

• “If obtain a pair not included, throw again.”
• “make the two dice distinguishable” (because 1,3 and 3,1 have different students) or “throw in order”

Condone extending to the entire sample space (e.g. “assign a further 2 pairs to each student and the remaining 3 to ‘throw again’”) but the method must remain random.

(b)(ii) B1: Accept \(\frac{1}{36}\)

Notes

M1: Attempting this calculation, allow wrong values but for this mark must be a fraction with a product in the numerator and a sum of two products in the denominator.

A1: Fully correct expression

A1: Or 133/159 or 0.8365 (4sf) (0.836477…)

Notes

M1: Setting up an expression in this form using the given values

M1: Setting their expression = 0.43 and attempting to solve

A1: cao (watch for 0.389 from incorrect working)

B1FT: “Around 38.8 or 39 or 40” (oe e.g. 2/5).
Must be in context and include "about" or "approximately" or "roughly" oe

Notes

B1: or showing 3, 14, 2 and 6 correctly placed
Allow omission of rectangle, so long as 14 seen outside
Allow probabilities in the diagram

M1: May be implied, eg by 2 seen in correct place in diagram

M1: Their \(x + 3\). May be implied by answer

If \(x\) is total English, giving \(x = 5\), use an equivalent scheme.

Alternative (incorrect) method for H \(\leftrightarrow\) E

M1: Or implied in diagram, History only = 2, or total History = 5

M1: Their \(3x + 3\). May be implied by answer

A1: If \(x\) is total History, giving \(x = 5\), use an equivalent scheme

Notes

M1: Allow omit \(\times 2\). Allow \(\dfrac{5}{25} \times \dfrac{20}{25}\) or 0.16 or 0.32. Allow + ….

A1: NB No ft from (a) in (b)

M2: M1 for one of \(\frac{6}{25} \times \frac{2}{24}\) or \(\frac{3}{25} \times \frac{14}{24}\)   oe
OR \(\frac{6}{25} \times \frac{2}{25} + \frac{3}{25} \times \frac{14}{25}\) (both terms)
OR \(\frac{6}{25} \times \frac{a}{24} + \frac{3}{25} \times \frac{b}{24}\) or \(\frac{2}{25} \times \frac{a}{24} + \frac{14}{25} \times \frac{b}{24}\) (\(a\), \(b\) integer \(\lt 24\))
Allow any of the above + extras for M1

M1: Divide attempted probs of correct events dep \(\geqslant\) M1M1

A1: Careful!! SCs for correct answer by incorrect methods:
“×2” omitted throughout:
\(\dfrac{9}{100} \div \dfrac{1}{6} = \dfrac{27}{50}\): M1A0M1M0M1A1 (Total 4)
Denominator 25×25 instead of 25×24:
\(\dfrac{54}{625} \div \dfrac{4}{25} = \dfrac{27}{50}\): M1A0M1M0M1A1 (Total 4)
Both the above \(\dfrac{27}{625} \div \dfrac{2}{25} = \dfrac{27}{50}\): M1A0M1M0M1A1

Alternative method 1

M1M1: M1 for one of 2×6 or 3×14
or M1 for \(2 \times a + 3 \times b\)   (\(a\), \(b\) integers, \(a \lt 23\), \(b \lt 22\))

Alternative method 2

M1: M1 for three of these fractions seen

A1: A1 for all four fractions seen

M3: M2 for one of these products \(6/20 \times 2/5\) or \(3/5 \times 14/20\)
or M1 for \(a/20 \times 2/5 + 3/5 \times b/20\)
or M1 for \(6/a \times 2/5 + 3/5 \times 14/b\)

Alternative (incorrect) method for H \(\leftrightarrow\) E

M1: Allow without ×2. Allow \(\dfrac{9}{25} \times \dfrac{16}{25}\) or 0.230 or 0.461.

M1: Attempt divide attempted probabilities of correct events dep at least M1M1

A0: SC answer \(\dfrac{3}{4}\), but omit ×2 and/or denominator of 25, M1A0M1M0M1A1

Notes

M1: Correct equation involving multiple of \(k\)

A1: Must see previous line and answer

Alternative

Notes

or equivalent exact values

Notes

M1: At least three of these seen or implied. No extras or repeats.

M1: At least two correct terms, no incorrect coefficients; ft their table.
Allow in terms of \(k\)

A1: Allow 0.0922 (3 sf)

Notes

B1B1 for both sets in any order, without extras.   Both soi.
B1 for both sets in any order, with extras.

M1: \(\left(\frac{12}{25}\right)^4 \times \frac{4}{25}\) or \(\left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2\) oe seen. Ignore coeffs. ft their table

A1: For either \(\left(\frac{12}{25}\right)^4 \times \frac{4}{25} \times 5\) or \(\left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2 \times {}^5\mathrm{C}_2\) oe ft their table

A1: Allow 0.106 (3 sf)

Alternative method for M1A1

M1: oe

A1: For either \(\left(\frac{12}{25}\right)^4 \times \frac{4}{25} \times (4 + 1)\) or \(\left(\frac{12}{25}\right)^3 \times \left(\frac{6}{25}\right)^2 \times ({}^4\mathrm{C}_2 + 4)\) oe

Notes

Allow 0%

Notes

Ignore extra branches if no probabilities or \(p = 0\)

B2: 8 correct branches and probs OR names, no extra branches
B2: 7 correct branches, probs and names, no extra branches

B1: 8 correct branches without probs & names. No extra branches
B1: 6 correct branches, probs and names. Ignore extra branches
Ignore products at ends

Notes

All correct   M2   ft their diagram
Two products correct M1   ft their diagram

SC Correct answer with no working: B2

Notes

M1: or \(1 - (0.7^2 + 0.3^2)\) or \(1 - (0.49 + 0.09)\)   Condone missing brackets
or \(0.7 \times 0.3 \times 0.7 + 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.7 + 0.3 \times 0.7 \times 0.3\)   oe
or \(2 \times 0.147 + 2 \times 0.063\)   oe Wholly correct method ft their diagram

A1: SC Correct answer with no working: B1

Notes

ft their diagram for M-marks

M2: soi   M1 for one correct product or 0.063

M1: Must attempt \(\dfrac{\text{P(B wins \& 3 points)}}{\text{Their (c) NOT (d) or } 0.3 \times 0.3 + 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.3}\)

A1: Allow 0.583 (3 sf)    SC no working \(\frac{7}{12}\) B2
But minimal working is OK, eg \(\dfrac{2 \times 0.063}{0.216} = \dfrac{7}{12}\)   M2M1A1

Notes

B1: 2 probabilities correctly placed

B1: all three correctly placed; ignore other values

Notes

M1: value for \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})\) found; allow M1 for 0.0418 unsupported;
allow if seen in part (a) on diagram and nowhere else

M1: may be implied by 0.38 or 0.11 even if wrongly attributed

A1: both correct; allow unsimplified

A1: allow \((0.3382 + 0.0418)\times(0.0682 + 0.0418)\); allow P(A) \(\times\) P(B) = 0.0418 if P(A) = 0.38 and P(B) = 0.11 seen elsewhere

Alternative method

M1: value for \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})\) found; allow M1 for 0.0418 unsupported
allow if seen in part (a) on diagram

M1: may be implied by 0.11
or may be implied by 0.38

A1: or \(\mathrm{P}(B|A) = \dfrac{0.0418}{(0.3382 + 0.0418)} = 0.11\); allow eg \(\mathrm{P}(A|B) = \dfrac{\mathrm{P}(\mathrm{A}\cap\mathrm{B})}{\mathrm{P}(\mathrm{B})} = 0.38\) if P(B) = 0.11 and \(\mathrm{P}(\mathrm{A}\cap\mathrm{B}) = 0.0418\) seen elsewhere

A1: or \(\mathrm{P}(B|A) = 0.11 = \mathrm{P}(B)\) so independent

Notes

B1: allow \(\mathrm{P}(\mathrm{A} \cap \mathrm{B}) =\) their 0.0418 so not mutually exclusive; must refer to probability;
do not allow
eg they have an intersection;
eg the intersection is 0.0418

Notes

B1: entries in bold

B1: 0.46

Notes

M1: must use values from their table; \(0 \lt\) their \(0.46 \lt\) their \(0.58 \lt 1\)

Notes

M1: condone omission of 2
allow recovery from bracket error

A1: mark the final answer
allow SC1 for correct answer unsupported

Notes

M1: allow 0.09 embedded in correct place in Venn diagram or contingency table; allow M1 for 9%

A1: allow SC1 for correct answer unsupported

Notes

M1: M0 for 0.12 from wrong working

A1: allow SC1 for correct answer unsupported

Notes

A1: if M0 allow SCB1 for \(0.16 \times 0.75 \neq 0.09\) so not independent

Notes

M1: Venn diagram with 2 overlapping regions and 0.22 correctly placed; condone incorrect or no labelling

A1: all probabilities or percentages correctly placed and correctly labelled; ignore values in intersection
allow if no box drawn
if labels are eg \(A\) and \(B\), \(A\) and \(B\) need to be defined

Notes

M1: may be implied by correct answer or by 24%

A1: do not allow 24%

Notes

M1: M0 if binomial distribution or sampling with replacement used

M1: dependent on award of first M1

Alternative

Notes

M1: condone omission of \(\times 4\) in first term;
M0 if binomial distribution or sampling with replacement used

M1: for addition of their terms; dependent on award of first M1

Alternative

M1: for \({}_{4}\mathrm{C}_{3} \times {}_{5}\mathrm{C}_{1}\)

M1: for addition of their terms; dependent on award of first M1

NB \(1 + 4 \times 5 = 21\)

Alternative

M1: for two of these terms

M1: for 1 – the sum of their 3 terms

NB \(\dfrac{2520}{3024} = \dfrac{5}{6}\)

Notes

M1: M0 if binomial distribution or sampling with replacement used

Alternative

M1: for either term

Notes

B1: FT their probabilities
no FT if binomial distribution or sampling with replacement used

Alternatively

FT their calculated probabilities from first three parts

Notes

M1: or M1 for probabilities in bold correct in table or marked correctly on Venn diagram

NB 0.3 from \(0.6 \times 0.5\) does not score

Notes

M1: \(\dfrac{\text{their } \mathrm{P}(A \cap B)}{0.5}\)

Notes

B1: or \(0.6 \times 0.5 = 0.3\) or \(\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B)\)
FT their values with correct argument