AQA Paper 3 2021 June — Question 14 7 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2021
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProbability Definitions
TypeCombined event algebra
DifficultyStandard +0.3 This is a standard probability question requiring systematic use of probability laws (complement rule, addition rule) and solving simultaneous equations. While it involves multiple steps and careful algebraic manipulation, the techniques are routine for A-level students who have practiced probability problems. The independence check at the end is straightforward once P(A) and P(B) are found.
Spec2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles
\(A\) and \(B\) are two events such that $$P(A \cap B) = 0.1$$ $$P(A' \cap B') = 0.2$$ $$P(B) = 2P(A)$$
  1. Find \(P(A)\) [4 marks]
  2. Find \(P(B|A)\) [2 marks]
  3. Determine if \(A\) and \(B\) are independent events. [1 mark]
Question 14:
AnswerMarks
14(a)Uses
P(A∪B)=P(A)+P(B)−P(A∩B)
with 0.1 substituted correctly
or
draws a Venn diagram with 0.1
AnswerMarks Guidance
in the correct region3.1a B1
P(A∪B)=0.8
0.8 = P(A) + 2P(A) – 0.1
3P(A) = 0.9
P(A) = 0.3
Uses P(A∪B)=0.8 in the
equation
PI by showing at least three of
0.2, 0.1, x – 0.1 or 2x – 0.1 in
the correct regions on the Venn
AnswerMarks Guidance
diagram1.1b B1
Substitutes for P(B) to form an
equation to find P(A)
PI by correct answer for P(A)
or
shows all of 0.2, 0.1, x – 0.1 and
2x – 0.1 in the correct regions
AnswerMarks Guidance
on the Venn diagram1.1a M1
Obtains P(A) = 0.31.1b A1
Subtotal4
QMarking Instructions AO
AnswerMarks
14(b)Uses conditional probability
formula with 0.1 and their P(A)
AnswerMarks Guidance
substituted correctly3.1a M1
P(A∩B)
AnswerMarks
P(BA)=
P(A)
0.1
=
0.3
1
=
3
Obtains correct answer
FT their P(A)
if 0.1 < P(A) < 1
Allow but not 0.33(…) for
1
AnswerMarks Guidance
̇1.1b A1F
0.3
AnswerMarks Guidance
Subto3tal2
QMarking Instructions AO
AnswerMarks
14(c)Deduces that A and B are not
independent by comparing with
0.1
or
AnswerMarks Guidance
showsP(B)=0.6≠ P(BA) 2.2a
P(A)×P(B)=0.3×0.6=0.18
≠ P(A∩B)
because P(AՈB) = 0.1
AnswerMarks Guidance
Subtotal1
Question Total7
QMarking Instructions AO 
Question 14:
--- 14(a) ---
14(a) | Uses
P(A∪B)=P(A)+P(B)−P(A∩B)
with 0.1 substituted correctly
or
draws a Venn diagram with 0.1
in the correct region | 3.1a | B1 | P(A∪B)=P(A)+P(B)−P(A∩B)
P(A∪B)=0.8
0.8 = P(A) + 2P(A) – 0.1
3P(A) = 0.9
P(A) = 0.3
Uses P(A∪B)=0.8 in the
equation
PI by showing at least three of
0.2, 0.1, x – 0.1 or 2x – 0.1 in
the correct regions on the Venn
diagram | 1.1b | B1
Substitutes for P(B) to form an
equation to find P(A)
PI by correct answer for P(A)
or
shows all of 0.2, 0.1, x – 0.1 and
2x – 0.1 in the correct regions
on the Venn diagram | 1.1a | M1
Obtains P(A) = 0.3 | 1.1b | A1
Subtotal | 4
Q | Marking Instructions | AO | Marks | Typical Solution
--- 14(b) ---
14(b) | Uses conditional probability
formula with 0.1 and their P(A)
substituted correctly | 3.1a | M1 | P(A∩B)= P(A)×P(B|A)
P(A∩B)
P(B|A)=
P(A)
0.1
=
0.3
1
=
3
Obtains correct answer
FT their P(A)
if 0.1 < P(A) < 1
Allow but not 0.33(…) for
1
̇ | 1.1b | A1F
0.3
Subto3tal | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 14(c) ---
14(c) | Deduces that A and B are not
independent by comparing with
0.1
or
showsP(B)=0.6≠ P(B| A) | 2.2a | R1 | Not independent as
P(A)×P(B)=0.3×0.6=0.18
≠ P(A∩B)
because P(AՈB) = 0.1
Subtotal | 1
Question Total | 7
Q | Marking Instructions | AO | Marks | Typical Solution
$A$ and $B$ are two events such that
$$P(A \cap B) = 0.1$$
$$P(A' \cap B') = 0.2$$
$$P(B) = 2P(A)$$

\begin{enumerate}[label=(\alph*)]
\item Find $P(A)$
[4 marks]

\item Find $P(B|A)$
[2 marks]

\item Determine if $A$ and $B$ are independent events.
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2021 Q14 [7]}}
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