A Level Mathematics 9709 Paper 53, Oct/Nov 2025

$[(0.6{)}^3=]0.216,\frac{27}{125}$

One term of form ${}^{11}{\text{C}}_x(p{)}^x(1-p{)}^{11-x}$ with $0<p<1,x\mathrm{\ne }0$ or $11$.

Correct un-simplified expression, no terms omitted leading to final answer (or no more than 6 'middle' terms omitted in Method 2).

$=0.545$

Expression of the form ${}^{16}{\text{C}}_x\times {}^{10}{\text{C}}_y$ with $x+y=6$ linked to a correct identified scenario.

Two identified outcomes evaluated accurately, accept un-simplified.

Sum of their values of 3 correct identified scenarios, no incorrect/repeated scenarios.

Total = 174 300

$\frac{7!}{2!}$ seen alone, not multiplied or divided in a calculation nor subtracted from another value (or $5!\times a,1⩽a⩽30$, 1 can be implied, $a$ is an integer. No other terms added or subtracted).

$\frac{l!}{m!}-6!,l=7,8\text{ or }9,m=1\text{ or }2$ (or $b!\times \frac{6\times 5}{2}$ or $b!\times {}^6{\text{C}}_2$ or $b!\times \frac{{}^6{\text{P}}_2}{2!}$, where $b=5,6\text{ or }7$).

$=1800$

Accept $\frac{7!}{2!}\times 4$ or 10080 (or $7!\times 4\times 2!$ or 40320 in Method 2) alone or as the numerator or denominator of a fraction.

Accept 90720 (or 362880 in Method 2) as denominator in a fraction.

$[\text{Probability}=]\frac{10080}{90720},\frac{1}{9},0.111$ (or $[\text{Probability}=]\frac{40320}{362880},\frac{1}{9},0.111$)

Table with correct $x$ values and one correct probability.

2 more probabilities correct linked to the correct $x$ value, need not be in table, accept un-simplified.

5 correct probabilities linked with correct outcomes.

Accept un-simplified or bold expression with probabilities ($0<p<1$) that add to 1. May be calculated in variance. FT their table with 4 or 5 probabilities adding to 1.

Appropriate variance formula using their $(\text{E}(X){)}^2$ value. FT their table with 4 or more probabilities ($0<p<1$) which need not sum to 1 or with an expression no more evaluated than shown.

$\frac{127}{64},1\frac{63}{64},1.98$

$1-{\left(\frac{3}{4}\right)}^n$ with $n=6,7,8$ (or Method 2 sum of geometric terms).

$=0.867$

Correct tree diagram structure. (3 tiers).

Counter and first marble probabilities correct with outcomes labelled.

Second marble probabilities correct with outcomes labelled.

At least 2 correct 3-term products linked to correct scenarios. May be clearly identified on tree diagram. FT from a 3-tier tree.

4 correct identified scenarios added. May be clearly identified on the tree diagram, condone identification by correct (or FT) un-simplified products.

$=\frac{39}{70},0.557$

Correct formula for conditional probability using *their* probabilities from 3-tier tree with *their* (b) $<1$ or correct in denominator.

$=\frac{8}{13},0.615$

Correct stem, ignore extra values (not in reverse, not split). If a split stem-and-leaf plot is used (i.e. stem values are repeated) the remaining B marks are available.

Correct Linnets labelled on left, leaves in order from right to left and lined up vertically, no commas or other punctuation.

Correct Puffins labelled on same diagram, leaves in order and lined up vertically, no commas or other punctuation.

Correct non-symmetrical key, for their diagram, need both team names and 'mins' at least once here or in leaf headings (or title).

$64⩽\text{UQ}⩽68-51⩽\text{LQ}⩽57$. Implied if both quartile values are stated and an appropriate IQR calculated accurately.

$=11$

Single linear scale from 45 to 75, at least 3 equally spaces values marked, using a scale of 2cm = 10 or 20 minutes only. Label to include time and minutes/min/m.

Correct plot labelled 'Linnets' with 5 key values using their linear scale. Daylight rule applied. Whiskers not at top/bottom of box or through box (penalise only once). Condone missing vertical lines on max/min values. LQ and UQ correct or FT their values from 6(b).

Correct plot labelled 'Puffins' with 5 key values using the same linear scale as Linnets. Whiskers not at top/bottom of box or through box (penalise only once). Condone missing vertical lines on max/min values.

Acceptable comment about spread/central tendency

Use of $\pm$ standardisation formula with 15.6, 14.8 and 1.5 substituted, condone ${\sigma }^2$, condone $\sqrt{\sigma }$, condone continuity correction of $\pm 0.05$.

Calculating the appropriate probability area (leading to their final answer, expect < 0.5).

$=0.297$

$-1.282$ or 1.282 seen, CAO – critical value.

Using $\pm$ standardisation formula with 14.8, 1.5 and $t$ equating to a $z$-value (not 0.8159, 0.1841, 0.9, 0.1, 0.5398, 0.4602 …) not ${\sigma }^2$, not $\sqrt{\sigma }$, no continuity correction.

$t=12.9$

Substituting 14.8 and 1.5 and either 13.8 or 15.8 appropriately into one $\pm$standardising formula, not ${\sigma }^2$ or $\sqrt{\sigma }$, no continuity correction.

Calculating the appropriate probability area (leading to their final answer, expect < 0.5) from their Z-values.

$=0.495(4)$

[Expected number of days =] $0.4954\times 5\times 45=111.465$, so 111 days