| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question using a tree diagram with clearly stated percentages. Parts (a)-(c) involve basic probability calculations (multiplication and addition along branches), while part (d) requires Bayes' theorem, but in a standard textbook format with no conceptual challenges. The question is easier than average A-level maths as it's purely procedural with no problem-solving insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks |
|---|---|
| 4(a) | B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | 0.3"0.98" | M1 |
| = 0.294 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | (0.30.02)+("0.45""0.04")+("0.25""0.06") | M1 |
| = 0.039 | A1 |
| Answer | Marks |
|---|---|
| (d) | "0.25""0.06" 0.015 |
| Answer | Marks |
|---|---|
| "0.039" "0.039" | M1,M1 |
| Answer | Marks |
|---|---|
| 13 | A1 |
| Answer | Marks |
|---|---|
| Notes | Total 9 |
| (a) | B1 for 0.45, 0.25 and 0.98 Allow fractions |
| Answer | Marks |
|---|---|
| (b) | M1 may ft their tree diagram if method shown 0.3 × " their 0.98" |
| Answer | Marks |
|---|---|
| (c) | M1 may ft their tree diagram if method shown |
| Answer | Marks |
|---|---|
| (d) | p p |
| Answer | Marks | Guidance |
|---|---|---|
| Qu | Scheme | Marks |
Question 4:
--- 4(a) ---
4(a) | B1B1
(2)
(b) | 0.3"0.98" | M1
= 0.294 | A1
(2)
(c) | (0.30.02)+("0.45""0.04")+("0.25""0.06") | M1
= 0.039 | A1
(2)
(d) | "0.25""0.06" 0.015
( )
P C Red = =
"0.039" "0.039" | M1,M1
5
= 0.3846… or
13 | A1
(3)
Notes | Total 9
(a) | B1 for 0.45, 0.25 and 0.98 Allow fractions
B1 0.04, 0.96 and 0.06, 0.94 Allow fractions
(b) | M1 may ft their tree diagram if method shown 0.3 × " their 0.98"
A1 0.294 oe
(c) | M1 may ft their tree diagram if method shown
A1 0.039 oe
(d) | p p
M1 allow or where 0 < p < 1 and p < denominator and their (c) is a
"their part (c)" 0.039
probability or
"0.25""0.06" 0.015
allow or where 0 < q < 1 and q > numerator
q q
"0.25""0.06"
M1 for ft their tree diagram and their part(c) if all 3 figures shown in
"0.039"
working. We will condone num > denom
A1 awrt 0.385
NB if correct ft on numerator and denominator leads to “num” > “denom” then max score
is M0M1A0
Qu | Scheme | MarksA bag contains a large number of coloured counters. Each counter is labelled A, B or C
30% of the counters are labelled A
45% of the counters are labelled B
The rest of the counters are labelled C
It is known that
2% of the counters labelled A are red
4% of the counters labelled B are red
6% of the counters labelled C are red
One counter is selected at random from the bag.
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram on the opposite page to illustrate this information. [2]
\item Calculate the probability that the counter is labelled A and is not red. [2]
\item Calculate the probability that the counter is red. [2]
\item Given that the counter is red, find the probability that it is labelled C [3]
\end{enumerate}
\includegraphics{figure_3}
\hfill \mbox{\textit{Edexcel S1 2023 Q4 [9]}}