A Level Mathematics 9709 Paper 53, May/June 2026

$[{\left(\frac{7}{8}\right)}^6\times \frac{1}{8}]=0.0561,\frac{117649}{2097152}$

$1-{\left(\frac{7}{8}\right)}^9$ or full addition of terms up to power 8

= 0.699

One term of the form ${}^{12}{\text{C}}_x(p{)}^x(1-p{)}^{12-x}$, $0<p<1$, $x\mathrm{\ne }0$ or 12.

Correct un-simplified expression.

= 0.746

$[22k=1]k=\frac{1}{22}$

Table with correct values of $x$ and at least one non-zero probability correct or FT their $k$, $0<k<1$.

4 correct probabilities linked with the correct outcomes.

$[\text{E}(X)=]-2\times k+[0+]1\times 4k+3\times 16k$

$\text{Var}(X)=\frac{1\times (-2{)}^2+[1\times 0^2+]4\times 1^2+16\times 3^2}{22}-(\text{their E}(X){)}^2$

= 1.74

At least 4 frequency densities calculated.

All bar heights correct on graph, NOT FT, daylight rule, using their suitable linear scale with at least 3 values indicated. At least 50% of the grid must be used for each axis.

Bar ends at 10, 15, 20, 25, 35, 60. All 6 bars drawn, with at least 3 values indicated, $0⩽\text{horizontal scale}⩽60$. At least 50% of the grid must be used for each axis.

Axes labelled frequency density (fd) time (t) and minutes, OE, or an appropriate title. (Axes may be reversed).

Midpoints: 5 12.5 17.5 22.5 30 47.5

Correct un-simplified mean formula with their 6 midpoints (not upper bound, lower bound, upper limit, lower limit, class width, frequency density, frequency and must be within class), accept un-simplified. Condone 1 value error on numerator.

$=22.1,22\frac{1}{15},22\frac{10}{150}$

$15⩽t<20$

Fully correct structure labelled tree diagram. Probabilities for coin correct $\frac{1}{4},\frac{3}{4}$.

Probabilities for Bag A and Bag B correct (irrespective of correct coin toss being present). First draw $\frac{5}{8},\frac{3}{8},\frac{4}{9},\frac{5}{9}$. Second draw $\frac{4}{7},\frac{3}{7},\frac{5}{7},\frac{2}{7},\frac{3}{8},\frac{5}{8},\frac{4}{8},\frac{4}{8}$.

$\frac{a}{4}\times \frac{b}{8}\times \frac{c}{7\text{ or }8}\text{ or }\frac{d}{4}\times \frac{e}{9}\times \frac{f}{8\text{ or }9}\text{ seen or }\frac{a}{4}\times \frac{b}{56\text{ or }64}\text{ or }\frac{c}{4}\times \frac{d}{72\text{ or }81}\text{ seen.}$

3 correct scenarios added – maybe identified on tree diagram or by seeing $\text{P}(\text{RR})+\text{P}(\text{BB})+\text{P}(\text{GG})$.

$\frac{151}{336},0.449$

$\text{P}(\text{H}\cap \text{1R or 2R})=\frac{1}{4}\times \frac{5}{8}\times \frac{4}{7}+\frac{1}{4}\times \frac{5}{8}\times \frac{3}{7}+\frac{1}{4}\times \frac{3}{8}\times \frac{5}{7}$

$\text{P}(\text{1R or 2R})=1-\text{P}(\text{BB})-\text{P}(\text{GG})$

Correct fraction calculation seen – accept un-simplified, or FT their clearly identified $\text{P}(\text{H}\cap \text{1R or 2R})$ and $\text{P}(\text{1R or 2R})$ used to form conditional probability.

$\frac{75}{257},0.292$

Use of $\pm$standardisation formula once with 125, 8 and either 105 or 135. No continuity correction. No ${\sigma }^2,\sqrt{\sigma }$.

Calculating the appropriate probability area, must be a probability. Values given here come from tables and may be slightly different if using values from a calculator. Leading to their final answer, expect > 0.5.

= 0.888

$0.8775⩽z<0.8785$, $-0.8785<z⩽-0.8775$ seen.

2.7 and 2.5 substituted in $\pm$ standardisation formula equated to a $z$-value, $(\pm \frac{0.2}{\sigma })$ equated to a $z$-value scores M1 BOD. No continuity correction, not ${\sigma }^2,\sqrt{\sigma }$.

$[\sigma =]\frac{100}{439},0.228$

64.8 and 12.312 or 12.31 seen, allow un-simplified. May be in standardisation formula. $(\sigma =\sqrt{12.312},3.5088,\frac{9\sqrt{95}}{25}\dots \text{to at least 4SF implies correct variance})$.

Substituting their 64.8 and their 12.312 into $\pm$standardisation formula, any number for 59.5, allow ${\sigma }^2,\sqrt{\sigma }$.

Use continuity correct 59.5 or 60.5 in their standardisation formula.

Appropriate area $\mathrm{\Phi }$, from final process, must be a probability. May be implied by a sketch of the required probability areas. Note: correct final answer implies this M1. Expect final answer < 0.5.

= 0.0655

Method 1: $a-6!,a>720$. Or $a-4!\times 5\times 6,a>720$, $a$ must be an integer. Method 2: $5!\times m,m>1$.

Method 1: $\frac{7!}{c!}-(k!\times n,n=1\text{ or }2),k=5\text{ or }6\text{ and }c=1\text{ or }2$. Method 2: $n\times {}^6{\text{C}}_2$ or $n\times {}^6{\text{C}}_4$ or $n\times \frac{6\times 5}{2}$ or $n\times \frac{{}^6{\text{P}}_2}{2!}$. Condone $n\times {}^6{\text{P}}_2$ or $n\times {}^6{\text{P}}_4$ or $n\times 6\times 5$, $n$ must be an integer.

= 1800

Method 1: $\frac{6!}{4!}\times b,b⩾1,b\text{ an integer.}$ Method 2: $\left(\frac{{}^6{\text{P}}_4}{4!}\text{ or }{}^6{\text{C}}_4\right)\times d,d⩾1,d\text{ an integer.}$

Method 1: $c\times 3,c⩾6,c\text{ an integer.}$ Method 2: $e\times 3!,e⩾6,e\text{ an integer.}$

= 90

One identified outcome ${}^4{\text{C}}_k\times n,k=1,2,3\text{ and }n⩾1$.

3 outcomes of the form ${}^4{\text{C}}_k\times n,k=1,2\text{ and }n⩾1$ un-simplified.

5 correct scenarios added (values do not need to be correct) may be identified by un-simplified expressions.

[Number of selections = ] 16