A Level Mathematics 9709 Paper 53, May/June 2025

4960 – 40k = 836 or $\frac{4960}{40}-k=\frac{836}{40}$

$k=103.1$

$[{\text{sd}}^2=]\frac{25410.8}{40}-{\left(\frac{836}{40}\right)}^2$ or $[{\text{sd}}^2=]\frac{622978.4}{40}-{124}^2$

$[{\text{sd}}^2=198.46,]$ sd = 14.1

$1-(0.55{)}^3$ or ${}^3{\text{C}}_1(0.45)(0.55{)}^2{+}^3{\text{C}}_2(0.45{)}^2(0.55)+(0.45{)}^3[0.55{]}^0$

$=0.834,\frac{6669}{8000}$

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

Correct un-simplified expression.

$=0.0996$

$[\text{P}(X=8)=]\left(\frac{1}{6}\right){\left(\frac{5}{6}\right)}^7=0.0465,\frac{78125}{1679616}$

$1-{\left(\frac{5}{6}\right)}^d,d=8,9$ or full addition of terms

$=0.767$

Two identified scenarios correct, accept un-simplified calculations (or 1 – 1 identified scenario correct in Method 2).

Four correctly identified scenarios added (or 1 – 2 correctly identified scenarios in Method 2). Condone ${}^2{\text{C}}_1=2$, ${}^5{\text{C}}_1=5$ etc.

$0.196,\frac{763}{3888}$

Accept 43 or 44

Median = 4.8 plotted on box.

LQ = 3.7 or 3.8, UQ = 6.5 or 6.6 plotted on box.

Correct end points for whiskers and whiskers drawn

Linear scale and labelled Time in minutes

Correct structure and probabilities for Bag A branches which must be the first set of branches, including labels R, B, G.

Correct structure and probabilities for one Bag B branch, including labels R, B, [G]. Condone inclusion of branches with 0 probability.

Correct structure and probabilities for remaining Bag B branches, including labels R, B, [G]. Condone inclusion of branches with 0 probability.

$[\text{P}(\text{RR})+\text{P}(\text{BB})+\text{P}(\text{GG})=]\frac{6}{12}\times \frac{6}{9}+\frac{5}{12}\times \frac{4}{9}+\frac{1}{12}\times \frac{1}{9}$

$=\frac{57}{108},\frac{19}{36}$

$\frac{5}{12}\times \frac{4}{9}$ or $\frac{20}{108}$ or $\frac{5}{27}$ or their P(BB) from Q5(b) seen as a numerator of a single fraction. FT tree diagram with values from only 2 bags.

$\frac{6}{12}\times \frac{3}{9}+\frac{5}{12}\times \frac{4}{9}+\frac{1}{12}\times \frac{3}{9}$ seen as the denominator of a single fraction. $\frac{5}{27}$ or their P(BB) may be seen from Q5b in place of $\frac{5}{12}\times \frac{4}{9}$. FT tree diagram with values from only 2 bags.

$=\frac{20}{41}$

Substituting 2.65, 2.50 and 0.12 into the ±standardisation formula allow ${\sigma }^2$ or $\sqrt{\sigma }$. Condone continuity correction $\pm 0.005$.

$=0.894$

Seeing $2.50+1.65\times 0.12$ or 2.698 and 2.50 and 0.12 substituted into the ±standardisation formula with a direct attempt to find the probability area.

$=1-0.9505=0.0495$

$\text{Expected number}=0.0495\times 160=7.92\text{, so 7 or 8}$

$0.738<z_1<0.740$ or $-0.740<z_1<-0.738$ seen.

$0.150<z_2<0.152$ or $-0.152<z_2<-0.150$ seen.

Use of the ±standardisation formula once with $\mu$, $\sigma$ equating to a $z$-value (not 0.77, 0.23, 0.44, 0.56, 0.7794, 0.5910, 0.6700, 0.7123, 0.2206, 0.4090, 0.3300, 0.2877). No continuity correction, not ${\sigma }^2$, not $\sqrt{\sigma }$.

Solve 2 equations in $\mu$ and $\sigma$ using the elimination method, substitution method or other appropriate approach to obtain values for both $\mu$ and $\sigma$.

$\mu =1.37,\sigma =0.153$

$7!-k,k$ an integer $⩾1$ (or $5!\times p$ where $p$ an integer $>1$).

$m-6!\times n:n=1,2$ where $m$ an integer $>6!$ ($n=1$ can be implied) (or $q\times 6\times 5,q{\times }^6{\text{C}}_2,q{\times }^6{\text{P}}_2,q\times \frac{6\times 5}{2}$, where $q$ an integer $>1$).

$=3600$

$7!\times r$, where $r$ is an integer, $r>1$ or $4!\times s$, where $s$ is an integer, $s>1$.

$7!\times 4!\times t$, where $t$ is an integer, $t⩾1$, $t=1$ can be implied.

241920

${}^{11}{\text{C}}_v{\times }^{11-v}{\text{C}}_u\times {[}^{11-v-u}{\text{C}}_w]$, $u,v,w=6,3,2,u\mathrm{\ne }v\mathrm{\ne }w$.

$=4620$

group of 6: ${}^8{\text{C}}_3{\times }^5{\text{C}}_3\times {(}^2{\text{C}}_2)=560$ (560 seen, accept un-simplified; or $\frac{4}{33},0.1212\dots$ seen)

group of 3: ${}^8{\text{C}}_6({\times }^2{\text{C}}_2{\times }^3{\text{C}}_3)=28$ (28 seen accept un-simplified; or $\frac{1}{165},0.00606(06\dots )$ seen)

$\text{Probability they are in same group}=\frac{\text{their}(560+28)}{\text{their (c)}}$ or $\frac{\text{their}(560+28)}{4620}$ (or $\text{their }\frac{4}{33}+\text{their }\frac{1}{165}$)

$=\frac{588}{4620},\frac{7}{55},0.127$