| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Bayes with sampling without replacement |
| Difficulty | Moderate -0.3 This is a standard S1 conditional probability question with tree diagrams and algebraic manipulation. Part (a) requires basic probability calculations, (b) is a straightforward algebraic proof, (c) involves solving a simple equation, and (d) applies the definition of conditional probability. While it requires multiple steps and some algebraic facility, all techniques are routine for S1 level with no novel insights needed. |
| 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 | Guidance |
|---|---|---|
| (a) \(\frac{n}{2n+1}\) Red; \(\frac{n+1}{2n+1}\) Black; \(\frac{n}{2n+1}\) and \(\frac{n+1}{2n+1}\) in the correct places on the tree diagram | B1 | For \(\frac{n}{2n+1}\) and \(\frac{n+1}{2n+1}\) in the correct places on the tree diagram |
| \(\frac{n-1}{2n}\) Red; \(\frac{n+1}{2n}\) Black; \(\frac{n-1}{2n}\) and \(\frac{n+1}{2n}\) in the correct places on the tree diagram | B1 | For \(\frac{n-1}{2n}\) and \(\frac{n+1}{2n}\) in the correct places on the tree diagram |
| \(\frac{n}{2n}\) Red; \(\frac{n}{2n}\) Black; \(\frac{n}{2n}\) and \(\frac{n}{2n}\) in the correct places on the tree diagram | B1 | For \(\frac{n}{2n}\) and \(\frac{n}{2n}\) in the correct places on the tree diagram. Allow \(\frac{1}{2}\) for \(\frac{n}{2n}\) in both places |
| (b) \('\frac{n}{2n+1}' \times '\frac{n+1}{2n}' + '\frac{n+1}{2n+1}' \times '\frac{n}{2n}'\) | M1 | For use of P(Red) × P(Black) + P(Black) × P(Red) ft their tree diagram |
| \(= \frac{2n(n+1)}{2n(2n+1)} = \frac{n+1}{2n+1}\) * | A1* | Answer is given so no incorrect working can be seen. Must have at least one correct line of working between M1 and the given answer. |
| (c) \(\frac{n+1}{2n+1} = \frac{25}{49} \Rightarrow n = 24\) | M1 A1 | For solving to find \(n = 24\) |
| So 49 counters in the box | Cao | |
| (d) \('\frac{25}{49}' \times '\frac{24}{48}' = \frac{1}{2}\) | M1 A1 | Allow a correct ratio in terms of \(n\) e.g. \(\frac{2n+1}{2n} \times \frac{n}{2n}\) oe ft their tree diagram for the numerator \(\frac{n+1}{2n+1}\) or |
| For 0.5 (Working must be seen) |
**(a)** $\frac{n}{2n+1}$ Red; $\frac{n+1}{2n+1}$ Black; $\frac{n}{2n+1}$ and $\frac{n+1}{2n+1}$ in the correct places on the tree diagram | B1 | For $\frac{n}{2n+1}$ and $\frac{n+1}{2n+1}$ in the correct places on the tree diagram
$\frac{n-1}{2n}$ Red; $\frac{n+1}{2n}$ Black; $\frac{n-1}{2n}$ and $\frac{n+1}{2n}$ in the correct places on the tree diagram | B1 | For $\frac{n-1}{2n}$ and $\frac{n+1}{2n}$ in the correct places on the tree diagram
$\frac{n}{2n}$ Red; $\frac{n}{2n}$ Black; $\frac{n}{2n}$ and $\frac{n}{2n}$ in the correct places on the tree diagram | B1 | For $\frac{n}{2n}$ and $\frac{n}{2n}$ in the correct places on the tree diagram. Allow $\frac{1}{2}$ for $\frac{n}{2n}$ in both places
**(b)** $'\frac{n}{2n+1}' \times '\frac{n+1}{2n}' + '\frac{n+1}{2n+1}' \times '\frac{n}{2n}'$ | M1 | For use of P(Red) × P(Black) + P(Black) × P(Red) ft their tree diagram
$= \frac{2n(n+1)}{2n(2n+1)} = \frac{n+1}{2n+1}$ * | A1* | Answer is given so no incorrect working can be seen. Must have at least one correct line of working between M1 and the given answer.
**(c)** $\frac{n+1}{2n+1} = \frac{25}{49} \Rightarrow n = 24$ | M1 A1 | For solving to find $n = 24$
So 49 counters in the box | | Cao
**(d)** $'\frac{25}{49}' \times '\frac{24}{48}' = \frac{1}{2}$ | M1 A1 | Allow a correct ratio in terms of $n$ e.g. $\frac{2n+1}{2n} \times \frac{n}{2n}$ oe ft their tree diagram for the numerator $\frac{n+1}{2n+1}$ or
| | For 0.5 (Working must be seen)
**Total: 9 marks**\begin{enumerate}
\item A box contains only red counters and black counters.
\end{enumerate}
There are $n$ red counters and $n + 1$ black counters.\\
Two counters are selected at random, one at a time without replacement, from the box.\\
(a) Complete the tree diagram for this information. Give your probabilities in terms of $n$ where necessary.\\
\includegraphics[max width=\textwidth, alt={}, center]{fe416f2e-bc81-444b-a0ca-f0eae9a8b149-24_940_1180_591_413}\\
(b) Show that the probability that the two counters selected are different colours is
$$\frac { n + 1 } { 2 n + 1 }$$
The probability that the two counters selected are different colours is $\frac { 25 } { 49 }$\\
(c) Find the total number of counters in the box before any counters were selected.
Given that the two counters selected are different colours,\\
(d) find the probability that the 1st counter is black.
You must show your working.
\hfill \mbox{\textit{Edexcel S1 2024 Q7}}