A Level Mathematics 9709 Paper 52, May/June 2026

Product of form ${}^6{\text{C}}_x\times {}^9{\text{C}}_y$, $x+y=5$, $x,y\mathrm{\ne }0$. Or 5 term probability of form $\frac{a}{15}\times \frac{b}{14}\times \frac{c}{13}\times \frac{d}{12}\times \frac{e}{11}$, $a,b,c,d,e⩾1$ seen.

3 correct identified outcomes added, accept un-simplified. Or Total – 3 correct identified outcomes added, accept un-simplified. Or $\frac{a}{15}\times \frac{b}{14}\times \frac{c}{13}\times \frac{d}{12}\times \frac{e}{11}\times {}^5{\text{C}}_f$, $f=1,2,3,4$ or $\frac{a}{15}\times \frac{b}{15}\times \frac{c}{15}\times \frac{d}{15}\times \frac{e}{15}\times {}^5{\text{C}}_f$, $f=1,2,3,4$ seen, $a,b,c,d,e⩾1$.

[Total number of ways =] ${}^{15}{\text{C}}_5$ Seen, used as denominator in a probability. Or Probabilities of 3 identified correct scenarios added, may be identified by un-simplified expressions. Or 1 – Probabilities of 3 identified correct scenarios added, may be identified by un-simplified expressions.

[Probability =] $\frac{861}{3003}=\frac{41}{143}=0.287$

${}^{13}{\text{C}}_x\times {}^{13-x}{\text{C}}_y[\times {}^{13-x-y}{\text{C}}_{13-x-y}]\times k$, $x\mathrm{\ne }y\mathrm{\ne }0$, $k=1,6$ or ${}^{13}{\text{P}}_x\times {}^{13-x}{\text{P}}_y\times {}^{13-x-y}{\text{P}}_{13-x-y}$, $x\mathrm{\ne }y\mathrm{\ne }0$. Or $\frac{13!}{a!\times b!\times c!}$, $a+b+c=13$, $a\mathrm{\ne }b\mathrm{\ne }c\mathrm{\ne }0$.

60060

Table with correct values of $x$ and at least one identified non-zero probability correct. Condone extra $x$ values if probability stated as 0.

Two more correct non-zero probabilities linked with correct outcome (3 correct probabilities present). Accept probabilities not in table if clearly identified.

4 correct non-zero probabilities linked with correct outcome. Accept probabilities not in table if clearly identified.

$(1-p{)}^6\times p$, $p=\text{their P}(2\cup 3)\text{ or P}(2)\text{ or P}(3)$, un-simplified. No other terms present.

0.0428

Either $5(r{)}^4(s{)}^2$ or $6(r{)}^5(s{)}^2$, $0<r,s<1$ seen, similar terms must be considered combined, no incorrect form terms present.

$a{t}^4(1-t{)}^2+b{t}^5(1-t{)}^2$, $a,b⩾1$, $0<t<1$.

= 0.191

[$6\times 7\times 1=$] 42

One identified outcome correct. Or Three identified outcomes correct. The last digits can be considered individually and listed… Or Two identified outcomes correct (Method 2/3/4).

Two correct odd/even scenarios added. May be implied by un-simplified expression or 30+45. Or Six correct first digit scenarios added, may be implied by un-simplified expression or 10+15+10+15+10+15. Or Three correct last digit scenarios added, condone 3 for 3C1. May be implied by un-simplified expression or 25 + 25 + 25. Or Three correct last digit and three ‘1’ and last digit scenarios added, may be implied by un-simplified expression or 20 + 20 + 20 + 5 + 5 + 5. Or Four correctly fully identified scenarios added, may be implied by un-simplified expression or 6 + 24 + 18 + 27.

[Total number of numbers = ] 75

One term of the form ${}^9{\text{C}}_x(p{)}^x(1-p{)}^{9-x}$, $0<p<1$, $x\mathrm{\ne }0$ or 9.

Correct un-simplified expression.

= 0.232

72 and 28.8 (CAO) seen, allow un-simplified. May be in standardisation formula. ($\sigma =\sqrt{28.8}$ or $\frac{12\sqrt{5}}{5}$ or 5.36656... to at least 4SF implies correct variance. Withold mark if variance clearly identified as standard deviation. Condone N(72, $\sqrt{28.8}$) if standardisation formula correct or variance/standard deviation stated correctly as well.

Substituting their µ and positive σ into the ± standardisation formula (any number for 65.5), allow ${\sigma }^2$.

Continuity correction 65.5 or 66.5 used in ±standardisation formula.

Appropriate area Φ, from final process, must be a probability. May be implied by a sketch of the required probability area.

= 0.113 final answer

$[\frac{5}{10}\times \frac{4}{10}=]\frac{1}{5}$

$\frac{a}{10}\times \frac{b}{10}+\frac{c}{10}\times \frac{d}{10}+e$, $a,b,c,d\mathrm{\ne }0$, $0⩽e<1$ or $\frac{a}{10}\times \frac{b}{11}+\frac{c}{10}\times \frac{d}{11}+e$, $a,b,c,d\mathrm{\ne }0$, $0⩽e<1$ or $\frac{a}{10}\times \frac{b}{9}+\frac{c}{10}\times \frac{d}{9}+e$, $a,b,c,d\mathrm{\ne }0$, $0⩽e<1$. May be implied by correct use of tree diagram. Correct fraction values may be simplified.

[P(R,R) + P(B,R) + P(G,R) =] $\frac{3}{10}\times \frac{b}{10}+\frac{5}{10}\times \frac{d}{10}+\frac{2}{10}\times \frac{g}{10}$, $b,d,g\mathrm{\ne }0$. or [P(R,R) + P(R',R) =]$\frac{3}{10}\times \frac{b}{10}+\frac{7}{10}\times \frac{d}{10}$, $b,d\mathrm{\ne }0$. May be implied by correct use of tree diagram. Probabilities need to be in a consistent order unless scenarios clearly stated. Correct fraction values may be simplified.

= $\frac{27}{100}$, 0.27[0]

$\frac{3}{10}\times \frac{8}{10}$ OE or 0.24 or $\frac{3}{10}\times \frac{8}{11}$ or $\frac{3}{10}\times \frac{8}{9}$ seen as numerator of fraction.

1 – their 5(b) or $\frac{3}{10}\times (\frac{5}{10}+\frac{3}{10})+\frac{5}{10}\times (\frac{4}{10}+\frac{3}{10})+\frac{2}{10}\times (\frac{5}{10}+\frac{2}{10})$ OE. or $\frac{3}{10}\times \left(\frac{8}{10}\right)+\frac{7}{10}\times \left(\frac{7}{10}\right)$ OE. Or 0.73. Seen as the denominator of fraction.

= $\frac{24}{73}$, 0.329

Correct stem, cannot be upside down, ignore extra values. If a split stem-and-leaf plot used (i.e. the stem values are repeated) the remaining B marks are available.

Correct Hippos labelled on left, leaves in order from right to left and lined up vertically (less than halfway to next column), no commas or other punctuation.

Correct Zebras labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas or other punctuation.

Correct non-symmetrical single key for their diagram, need both teams identified and ‘cm’ stated at least once here or in leaf headings or title.

188 ⩽ UQ ⩽ 191 – 168 ⩽ LQ ⩽ 176. Implied if both quartile values are stated and appropriate IQR calculated accurately.

= 18 (cm)

$15\times \frac{1}{2}$ or 7.5 seen. Or $\frac{2725}{15}\text{ or }\frac{545}{3}\text{ or }\frac{2541+x}{15}$ or (181.66... or 182.16...) to at least 4SF seen.

[$184+7.5=$] 191.5 [cm] or 192 [cm]

0.841 < z1 < 0.843 or −0.843 < z1 < −0.841 seen.

0.523 < z2 < 0.525 or −0.525 < z2 < −0.523 seen.

Use of the ± standardisation formula once with µ, σ and either 3.6 or 5.4, equated to a z-value (not 0.20, 0.80, 0.70, 0.30, 0.6179, 0.3821, 0.7580, 0.2420, 0.5793, 0.4207, 0.7881, 0.2119, 1 – z-value, 0.476, 0.158 etc). Not ${\sigma }^2$, $\sqrt{\sigma }$, allow continuity correction ±0.05.

Solve 2 equations in µ and σ with an attempt at the elimination method, substitution method or other appropriate approach to obtain values for both µ and σ.

µ = 8.37, σ = 5.66

Either use of ± standardisation formula once with 4.63, 0.55 and either 4.03 or 5.23, or use of ± standardisation formula once with 0.55 and either −0.6 or 0.6 as numerator. No ${\sigma }^2$, $\sqrt{\sigma }$. No continuity correction.

Calculating the appropriate probability area, must be a probability. May be implied by a sketch. (leading to their final answer. Expect > 0.5).

0.7246

[Expected number =] $0.7246\times 600$ (= 434.76) so 434