A Level Mathematics 9709 Paper 51, May/June 2026

Table with correct x values and at least 1 correct probability.

Three more correct probabilities correctly linked to the correct x values, need not be in table, accept un-simplified (4 correct probabilities now present).

6 correct probabilities linked with correct x values, need not be in table.

[E(X) =] $-1\times \frac{1}{16}+0\times \frac{3}{16}+1\times \frac{4}{16}+2\times \frac{5}{16}+3\times \frac{1}{16}+4\times \frac{2}{16}$

[Var(X) =] Appropriate variance formula using their $(\text{E}(X){)}^2$ value.

$=\frac{15}{8},1\frac{7}{8},1.875$

$[\text{P}(A\cup B)=]\frac{100}{150},\frac{2}{3},0.667$

$\text{P}(M)\times \text{P}(B)=\frac{60}{150}\times \frac{70}{150}$, $\text{P}(M\cap B)=\frac{28}{150}$ or Method 2: $\frac{\text{P}(B\cap M)}{\text{P}(M)}=\frac{28}{150}÷\frac{60}{150}$, $\text{P}(B\mathrm{/}M)=\frac{28}{60}$

$\text{P}(M\cap B)=\frac{14}{75},0.1867\text{P}(M)\times \text{P}(B)=\frac{14}{75},0.1867$ [events $M$ and $B$ are independent] or Method 2: evaluated with conclusion stated.

$[\text{P}(M\mathrm{\mid }A)=]\frac{14}{30},\frac{7}{15},0.467$

$[\text{P}(M^{\mathrm{\prime }}\cap B^{\mathrm{\prime }})=]\frac{48}{150},\frac{8}{25},0.32[0]$

Correct stem, ignore extra values (not in reverse, not split).

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

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

Correct non-symmetrical key, for their diagram, need both team names and ‘cm’ at least once here, or in leaf headings or a suitable title

Median = 21[cm]

$[\text{IQR}=]35-16$

$=19[\text{cm}]$

Medians very close, so average amount of snow is very similar

Spread of number of cm of snow more for Aldeny

$1-{\left(\frac{7}{12}\right)}^k$, for $k=5$ or $6$. (or Method 2 geometric series terms)

$=0.932$

$k{p}^2(1-p{)}^7$ or $k{p}^2(1-p{)}^8$, $k⩾1$, $k$ an integer $0<p<1$. No additional terms, may be in stages.

$8\times {\left(\frac{5}{12}\right)}^2\times {(1-\frac{5}{12})}^7$ or $9\times {\left(\frac{5}{12}\right)}^2\times {(1-\frac{5}{12})}^8$ seen.

$\text{Sum}=0.0319(22)$

One term ${}^{10}{\text{C}}_x(p{)}^x(1-p{)}^{10-x}$. With $0<p<1,x\mathrm{\ne }0$ or $10$.

Correct un-simplified expression. Allow 10 for ${}^{10}{\text{C}}_1$ (or ${}^{10}{\text{C}}_{10}$).

$=0.142$

$\frac{d}{9}\times \frac{e}{8}\times \frac{f}{7}$, $d<9,e<8,f<7$ seen.

One identified outcome case correctly evaluated. Accept un-simplified.

Three correct scenarios added. Un-simplified calculation can identify scenario.

Probability = $\frac{384}{504},\frac{16}{21}$ (or $\frac{64}{84},\frac{16}{21}$)

$\frac{7!}{3!}$ seen alone (not multiplied/divided) or Method 2: $4!\times k$, where $k⩾1$, and $k$ an integer.

$m-5!,m>120$ or Method 2: $q\times {}^5{\text{C}}_3\times 3$, where $q⩾1$, and $k$ an integer.

720

$\frac{7!}{3!}\times i,i⩾1$ where $i$ an integer or Method 2: ${}^7{\text{P}}_3\times k,k⩾1$ where $k$ an integer.

$\frac{7!}{j!}\times 5$, where $j=1,2,3$ or Method 2: ${}^7{\text{P}}_m\times \frac{5!}{3!}$ or ${}^7{\text{P}}_m\times 20$, where $m=2,3,4$ or ${}^7{\text{C}}_3\times \frac{5!}{3!}$ or ${}^7{\text{C}}_3\times 20$.

$=4200$

Substituting 2.74 and 0.25 and either 2.4 or 3.1 appropriately into one ±standardising formula, not ${\sigma }^2$ or $\sqrt{\sigma }$, no continuity correction.

Appropriate area $\mathrm{\Phi }$, from final process. Must be a probability.

$=0.8382$

$[\text{Expected number}=0.8382\times 250=209.55]$ $209$

$0.3055⩽z⩽0.3065$ or $-0.3065⩽z⩽-0.3055$.

±standardisation formula with 3.15, 0.28, $\mu$ equated to a $z$-value (not 0.62, 0.38, 0.7257, 0.2743, 0.6179, 0.3821, 0.7324, $1-\text{their }z\text{-value}$).

$\mu =3.24$

55.8 and 21.204 seen, allow un-simplified, may be seen in the standardisation formula.

Substituting their $\mu$ and positive $\sigma$ into the ± standardising formula (any number for 50.5), allow ${\sigma }^2$ or $\sqrt{\sigma }$.

Use continuity correction 49.5 or 50.5 in ±standardisation formula.

Appropriate area $\mathrm{\Phi }$, from final process, must be a probability.

$=0.875$