AQA Paper 3 Specimen — Question 15 6 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConditional Probability
TypeFinding unknown probability from total probability
DifficultyStandard +0.8 This is a multi-step conditional probability problem requiring students to set up and solve simultaneous equations using P(B|A), P(A|B), and the complement information. It demands careful algebraic manipulation of probability definitions (P(B|A) = P(A∩B)/P(A)) and systematic reasoning, going beyond routine conditional probability exercises but not requiring advanced techniques.
Spec2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles

A sample of 200 households was obtained from a small town. Each household was asked to complete a questionnaire about their purchases of takeaway food. \(A\) is the event that a household regularly purchases Indian takeaway food. \(B\) is the event that a household regularly purchases Chinese takeaway food. It was observed that \(P(B|A) = 0.25\) and \(P(A|B) = 0.1\) Of these households, 122 indicated that they did not regularly purchase Indian or Chinese takeaway food. A household is selected at random from those in the sample. Find the probability that the household regularly purchases both Indian and Chinese takeaway food. [6 marks]

Question 15:
AnswerMarks Guidance
15Uses conditional
probability, either  or AO3.1b M1
= 
P(A) 4
⇒P(A)=4P(A∩B)
P(A∩B) 1
= 
P(B) 10
⇒P(B)=10P(A∩B)
122 39
P(A∪B)=1− =
200 100
39
P(A)+P(B)−P(A∩B)= 
100
39
4P(A∩B)+10P(A∩B)−P(A∩B)=
100
3
P(A∩B)=
100
Obtains both equations
AnswerMarks Guidance
 and  correctlyAO1.1b A1
Evaluates P(A∪B)
AnswerMarks Guidance
correctly PIAO1.1b B1
Uses addition lawAO1.1a M1
Combines the three
AnswerMarks Guidance
equationsAO1.1a M1
Obtains correct
probability, as a fraction
AnswerMarks Guidance
or decimalAO2.2b A1
OR
AnswerMarks Guidance
ALTALT Produces a relevant
AO1.1bAO3.1b M1
A1M1 200
B A
9x x 3x
122
AnswerMarks
Venn diagram200
B A
9x x 3x
AnswerMarks Guidance
AO1.1bA1 122
Labels Venn diagram
correctly
AnswerMarks Guidance
Forms correct equationAO1.1b B1
to find x PI
AnswerMarks Guidance
Combines termsAO1.1a M1
Solves equationAO1.1a M1
Obtains correct
AnswerMarks Guidance
probabilityObtains correct AO2.2b
P(A∩B)= or 0.03
200
probability
AnswerMarks
Total6
TOTAL100
Question 15:
15 | Uses conditional
probability, either  or  | AO3.1b | M1 | P(A∩B) 1
= 
P(A) 4
⇒P(A)=4P(A∩B)
P(A∩B) 1
= 
P(B) 10
⇒P(B)=10P(A∩B)
122 39
P(A∪B)=1− =
200 100
39
P(A)+P(B)−P(A∩B)= 
100
39
4P(A∩B)+10P(A∩B)−P(A∩B)=
100
3
P(A∩B)=
100
Obtains both equations
 and  correctly | AO1.1b | A1
Evaluates P(A∪B)
correctly PI | AO1.1b | B1
Uses addition law | AO1.1a | M1
Combines the three
equations | AO1.1a | M1
Obtains correct
probability, as a fraction
or decimal | AO2.2b | A1
OR
ALT | ALT | Produces a relevant | AO3.1b
AO1.1b | AO3.1b | M1
A1 | M1 | 200
B A
9x x 3x
122
Venn diagram | 200
B A
9x x 3x
AO1.1b | A1 | 122
Labels Venn diagram
correctly
Forms correct equation | AO1.1b | B1 | 9x + x + 3x = 200 – 122
to find x PI
Combines terms | AO1.1a | M1 | 13x = 78
Solves equation | AO1.1a | M1 | x = 6
Obtains correct
probability | Obtains correct | AO2.2b | AO2.2b | A1 | A1 | 6
P(A∩B)= or 0.03
200
probability
Total | 6
TOTAL | 100
A sample of 200 households was obtained from a small town.

Each household was asked to complete a questionnaire about their purchases of takeaway food.

$A$ is the event that a household regularly purchases Indian takeaway food.

$B$ is the event that a household regularly purchases Chinese takeaway food.

It was observed that $P(B|A) = 0.25$ and $P(A|B) = 0.1$

Of these households, 122 indicated that they did not regularly purchase Indian or Chinese takeaway food.

A household is selected at random from those in the sample.

Find the probability that the household regularly purchases both Indian and Chinese takeaway food. [6 marks]

\hfill \mbox{\textit{AQA Paper 3  Q15 [6]}}