Question 6 - A-Level Maths

Edexcel S1 2023 June — Question 6 9 marks

Exam Board	Edexcel
Module	S1 (Statistics 1)
Year	2023
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Probability Definitions
Type	Combined event algebra
Difficulty	Moderate -0.3 This is a straightforward S1 probability question requiring standard formulas: P(A∪C) uses mutually exclusive property, P(B) uses P(A∪B) = P(A) + P(B) - P(A∩B) with conditional probability, and the Venn diagram applies independence systematically. All techniques are routine bookwork with clear signposting, making it slightly easier than average despite multiple parts.
Spec	2.03a Mutually exclusive and independent events 2.03b Probability diagrams: tree, Venn, sample space

Three events $A$, $B$ and $C$ are such that $$\mathrm{P}(A) = 0.1 \quad \mathrm{P}(B|A) = 0.3 \quad \mathrm{P}(A \cup B) = 0.25 \quad \mathrm{P}(C) = 0.5$$ Given that $A$ and $C$ are mutually exclusive

find P$(A \cup C)$ [1]
Show that P$(B) = 0.18$ [3]

Given also that $B$ and $C$ are independent,

draw a Venn diagram to represent the events $A$, $B$ and $C$ and the probabilities associated with each region. [5]

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
6(a)	0.6	B1

(1)

Answer	Marks
(b)	(AB)

P

  P (AB)=  0.10.3 or 0.3=

Answer	Marks
0.1	M1
0.25=0.1+P(B)−"0.03"or 0.25=0.1+P(B)−P(AB)	M1

(B)−0.15

P

0.25=0.1+P(B)−0.03or 0.3=  P(B) = 0.18*

Answer	Marks
0.1	A1*

(3)

Answer	Marks
(c)	M1

M1

B1ft

A1

(5)

Answer	Marks
Notes	Total 9
(a)	B1cao
(b)	(AB)

P

Answer	Marks
M1 for use of P (B	A)= with 0.1 and 0.3 substituted. Allow for 0.10.3 seen

P(A)

M1 0.25=0.1+P(B)− p where 0 p1 or p=P(AB) oe eg 0.25−0.1+ p=P(B) (allow

any letter for P(B))

A1* P(B) = 0.18 depends on both previous M marks for a fully correct equation in terms of P(B)

(allow any letter for P(B)) followed by P(B) = 0.18

NB 0.03 used/stated with no working could get M0M1A0

P(AB)=0.1P(B)then

Using they get M0M0A0

Verification could get M1M1A0

M1 for 0.10.3

0.18−0.15

M1 for 0.25−0.18−0.1=−0.03 or 0.3= or 0.25=0.1+0.18−P(AB)

0.1

Answer	Marks
(c)	M1 for 3 circles as per either diagram. If using Diagram 2 we must see exactly 2 zeros in one of the

intersections (as shaded). (Do Not accept blank or dash instead of 0) Condone missing rectangle.

Ignore labels

M1 for 0.09 and 0.41 marked correctly in diagram – condone incorrect/no label but must be in the

left or right hand circles in 1st diagram or must have zeros (condone blank or dash) in the 2 other

regions of the circle if in 2nd diagram

B1ft their "0.03" in correct place on diagram. Correct label required

B1ft for 0.34 or ft 0.75 – "their 0.41" where their 0.41 0.5 No other ft accepted. Do not allow

0.75 A1 fully correct Venn diagram including the rectangle and all 3 labels.

SC no labels could get M1M1B0B1A0 if using 3 intersecting circles must have blanks or 0 for the

2nd M1

Answer	Marks	Guidance
Qu	Scheme	Marks

Question 6:
--- 6(a) ---
6(a) | 0.6 | B1
(1)
(b) | (AB)
P
  P (AB)=  0.10.3 or 0.3=
0.1 | M1
0.25=0.1+P(B)−"0.03"or 0.25=0.1+P(B)−P(AB) | M1
(B)−0.15
P
0.25=0.1+P(B)−0.03or 0.3=  P(B) = 0.18*
0.1 | A1*
(3)
(c) | M1
M1
B1ft
B1ft
A1
(5)
Notes | Total 9
(a) | B1cao
(b) | (AB)
P
M1 for use of P (B| A)= with 0.1 and 0.3 substituted. Allow for 0.10.3 seen
P(A)
M1 0.25=0.1+P(B)− p where 0 p1 or p=P(AB) oe eg 0.25−0.1+ p=P(B) (allow
any letter for P(B))
A1* P(B) = 0.18 depends on both previous M marks for a fully correct equation in terms of P(B)
(allow any letter for P(B)) followed by P(B) = 0.18
NB 0.03 used/stated with no working could get M0M1A0
P(AB)=0.1P(B)then
Using they get M0M0A0
Verification could get M1M1A0
M1 for 0.10.3
0.18−0.15
M1 for 0.25−0.18−0.1=−0.03 or 0.3= or 0.25=0.1+0.18−P(AB)
0.1
(c) | M1 for 3 circles as per either diagram. If using Diagram 2 we must see exactly 2 zeros in one of the
intersections (as shaded). (Do Not accept blank or dash instead of 0) Condone missing rectangle.
Ignore labels
M1 for 0.09 and 0.41 marked correctly in diagram – condone incorrect/no label but must be in the
left or right hand circles in 1st diagram or must have zeros (condone blank or dash) in the 2 other
regions of the circle if in 2nd diagram
B1ft their "0.03" in correct place on diagram. Correct label required
B1ft for 0.34 or ft 0.75 – "their 0.41" where their 0.41 0.5 No other ft accepted. Do not allow
0.75
A1 fully correct Venn diagram including the rectangle and all 3 labels.
SC no labels could get M1M1B0B1A0 if using 3 intersecting circles must have blanks or 0 for the
2nd M1
Qu | Scheme | Marks

Show LaTeX source

Three events $A$, $B$ and $C$ are such that

$$\mathrm{P}(A) = 0.1 \quad \mathrm{P}(B|A) = 0.3 \quad \mathrm{P}(A \cup B) = 0.25 \quad \mathrm{P}(C) = 0.5$$

Given that $A$ and $C$ are mutually exclusive

\begin{enumerate}[label=(\alph*)]
\item find P$(A \cup C)$ [1]
\item Show that P$(B) = 0.18$ [3]
\end{enumerate}

Given also that $B$ and $C$ are independent,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item draw a Venn diagram to represent the events $A$, $B$ and $C$ and the probabilities associated with each region. [5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2023 Q6 [9]}}

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