A Level Mathematics 9709 Paper 51, Oct/Nov 2025

Using sum of probabilities =1 to form an equation in $k$ or value of $k$ stated.

Table with at least 2 correctly linked probabilities accurate. May be in terms of $k$.

Condone extra $X$ values if probability stated as 0.

4 correctly linked probabilities accurate. May not be in a table.

$[\text{E}(X)=28k=]\frac{14}{9},1.56$

$$\text{P}(X\mathrm{\ne }2\mid X>0)=\frac{\frac{10}{18}}{\frac{14}{18}}$$

$$=\frac{5}{7},0.714$$

$$[\frac{1}{6}{\left(\frac{5}{6}\right)}^7=]0.0465$$

${\left(\frac{5}{6}\right)}^4{\left(\frac{1}{6}\right)}^3\times d,d⩾1$.

$p^e(1-p{)}^{7-e}\times {}^6{\text{C}}_2,0<p<1$, accept $\frac{{}^6{\text{P}}_2}{2}$.

0.0335

Median = [$]32400

[IQR = $]33500\text{–}[$]31300

= [$]2200

$$[\text{Mean}=]\frac{878.3+896.5}{54}[=32.867]$$

so [$]32900

$$[\text{Variance}=][\frac{28616.09+29815.63}{54}-{\left(\frac{1774.8}{54}\right)}^2=]\frac{58431.72}{54}-{\left(\frac{1774.8}{54}\right)}^2$$

= 1.8511

$\sigma =1.36$, so $\sigma =[\mathrm{\$}]1360$

Bag B branches completed correctly including Red/Blue.

2 sets of Bag A branches completely correctly including Red/Blue.

Final 2 sets of Bag A branches completely correctly including Red/Blue. Penalise omission of Red/Blue only once. If not identified, assume RB continues as for first Bag A.

$$[\text{P}(\text{RRR})+\text{P}(\text{BBB})=]\frac{8}{11}\times \frac{4}{5}\times \frac{7}{10}+\frac{3}{11}\times \frac{2}{6}\times \frac{3}{11}$$

$$[\frac{224}{550}+\frac{18}{726}]=\frac{1307}{3025},0.432$$

One term of the form ${}^7{\text{C}}_x(p{)}^x(1-p{)}^{7-x}$. $0<p<1,x\mathrm{\ne }0$ or 7.

Correct un-simplified expression no terms omitted leading to the final answer. (Method 2: Correct un-simplified expression, no more than 2 ‘middle’ terms omitted leading to the final answer. Condone omission of final bracket ‘}’. If other brackets omitted, allow recovery if correct answer obtained.)

= 0.420

$$[\text{Mean}=0.1\times 150=]15$$ $$[\text{Variance}=0.1\times 150\times 0.9=]13.5$$

Substituting their $\mu$ and $\sigma$ into the $\pm$standardising formula (any number for 21.5), allow ${\sigma }^2$ or $\sqrt{\sigma }$.

Use continuity correction 21.5 or 22.5 in standardisation formula.

$[\text{P}(Z<1.769)=\mathrm{\Phi }(1.769)]$ Appropriate area $\mathrm{\Phi }$, from final process, must be a probability

0.962

$$\text{P}(\frac{148-155}{6}<Z<\frac{160-155}{6})$$

Calculating the appropriate probability area (leading to their final answer. Expect $p>0.5$).

= 0.6759

$[\text{Expected number}=0.6759\times 350=236.565]$ 236

$1.405<z_1<1.407,-1.407<z_1<-1.405$ seen.

$0.841<z<0.843,-0.843<z<-0.841$ seen.

Use of the $\pm$standardisation formula once with $\mu ,\sigma$ equating to a $z$-value (not 0.20, 0.80, 0.08, 0.92, 0.5793, 0.4207, 0.5319, 0.4681, 0.7881, 0.2119, 0.8212, 0.1788). Condone continuity correction $\pm 0.05$, not ${\sigma }^2$, not $\sqrt{\sigma }$.

Solve two equations in $\mu$ and $\sigma$ using the elimination method, substitution method or other appropriate approach to obtain values for both $\mu$ and $\sigma$.

$$\mu =105,\sigma =6.45$$

$$[\frac{10!}{2!2!3!}=]151200$$

$\frac{7!}{2!3!}\times k,1⩽k$, a positive integer. (Method 2: $\frac{6!}{2!3!}\times l,\frac{6!}{2!2!}\times m$ and $\frac{6!}{3!}\times n$ seen. $1⩽l,m,n$, positive integers. Method 3: $\frac{7!}{2!3!}\times l,\frac{7!}{2!2!}\times m$ and $\frac{7!}{2!}\times n$ seen. $1⩽l,m,n$, positive integers.)

Integer $\times 7$. (Method 2 & 3: Summing 4 correct or correctly identified scenarios, oe.)

= 5880

$\frac{8!}{2!3!}-q,1<q<3360$.

$m-\frac{6!}{2!},m>360$.

= 3000

[Number of ways with 3 Es =] ${}^7{\text{C}}_2(=21)$ (Method 2: $\frac{3}{11}\times \frac{2}{6}\times \frac{3}{11}$)

[Total number of ways is] ${}^{10}{\text{C}}_5(=252)$ (Method 2: ${}^5{\text{C}}_2\times k,0<k<1$. Accept ${}^5{\text{C}}_3\times k,0<k<1$.)

$[\text{Probability}=]\frac{21}{252},\frac{1}{12}$