| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram question with conditional probabilities. Parts (a)-(c) involve basic probability tree construction and multiplication/addition of probabilities. Parts (d)-(e) require conditional probability (Bayes' theorem), which is standard S1 content. The stopping condition adds minimal complexity since students just follow the branches. All techniques are routine for this specification level. |
| 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 |
|---|---|---|
| Tree diagram with correct probabilities | B1 | For 0.3 in the correct place on the first branch and 0.4 in the correct place on the second branch |
| B1 | For 0.2 and 0.8 in the correct place in the second branch | |
| B1 | For 0.2, 0.8, 0.6 and 0.4 in the correct place in the third branch. NB ISW any extra branches drawn on the tree diagram |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.7 \times 0.6 = 0.42\) oe | M1 | For \(0.7 \times 0.6\) |
| A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(`0.42` + (0.7 \times `0.4` \times `0.2`) + (`0.3` \times `0.2` \times `0.6`) = 0.512\) oe | M1 | For \(`0.42`+(0.7\times`0.4`\times`0.2`)+(`0.3`\times`0.2`\times`0.6`)\) Follow through part (b) and their tree diagram |
| A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{`0.42`}{`0.512`} = 0.8203\ldots\) oe, awrt 0.820 | M1 | For \(\dfrac{\text{part (b)}}{\text{part (c)}}\) provided the answer is a probability or ft their tree diagram |
| A1ft | awrt 0.820 or ft part (b) and part (c) provided the answer is a probability or ft their tree diagram. Allow 0.82 if ft and a decimal answer is given then this must be at least 3sf |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{`0.42`+(0.7\times`0.4`\times`0.2`)}{0.7} = 0.68\) oe or \(0.6+`0.4`\times`0.2` = 0.68\) oe | M1 | For a correct ratio of probabilities. Follow through their part (b) and their tree diagram or \(0.6+`0.4`\times`0.2`\) ft their tree diagram |
| A1 | Cao. Allow 0.680 |
**Question 1:**
**Part (a):**
Tree diagram with correct probabilities | B1 | For 0.3 in the correct place on the first branch and 0.4 in the correct place on the second branch
| B1 | For 0.2 and 0.8 in the correct place in the second branch
| B1 | For 0.2, 0.8, 0.6 and 0.4 in the correct place in the third branch. NB ISW any extra branches drawn on the tree diagram
**(3 marks)**
---
**Part (b):**
$0.7 \times 0.6 = 0.42$ oe | M1 | For $0.7 \times 0.6$
| A1 | Cao
**(2 marks)**
---
**Part (c):**
$`0.42` + (0.7 \times `0.4` \times `0.2`) + (`0.3` \times `0.2` \times `0.6`) = 0.512$ oe | M1 | For $`0.42`+(0.7\times`0.4`\times`0.2`)+(`0.3`\times`0.2`\times`0.6`)$ Follow through part (b) and their tree diagram
| A1 | Cao
**(2 marks)**
---
**Part (d):**
$\dfrac{`0.42`}{`0.512`} = 0.8203\ldots$ oe, awrt 0.820 | M1 | For $\dfrac{\text{part (b)}}{\text{part (c)}}$ provided the answer is a probability or ft their tree diagram
| A1ft | awrt 0.820 or ft part (b) and part (c) provided the answer is a probability or ft their tree diagram. Allow 0.82 if ft and a decimal answer is given then this must be at least 3sf
**(2 marks)**
---
**Part (e):**
$\dfrac{`0.42`+(0.7\times`0.4`\times`0.2`)}{0.7} = 0.68$ oe or $0.6+`0.4`\times`0.2` = 0.68$ oe | M1 | For a correct ratio of probabilities. Follow through their part (b) and their tree diagram or $0.6+`0.4`\times`0.2`$ ft their tree diagram
| A1 | Cao. Allow 0.680
**(2 marks)**
---
**Total: 11 marks**\begin{enumerate}
\item Sally plays a game in which she can either win or lose.
\end{enumerate}
A turn consists of up to 3 games. On each turn Sally plays the game up to 3 times. If she wins the first 2 games or loses the first 2 games, then she will not play the 3rd game.
\begin{itemize}
\item The probability that Sally wins the first game in a turn is 0.7
\item If Sally wins a game the probability that she wins the next game is 0.6
\item If Sally loses a game the probability that she wins the next game is 0.2\\
(a) Use this information to complete the tree diagram on page 3\\
(b) Find the probability that Sally wins the first 2 games in a turn.\\
(c) Find the probability that Sally wins exactly 2 games in a turn.
\end{itemize}
Given that Sally wins 2 games in a turn,\\
(d) find the probability that she won the first 2 games.
Given that Sally won the first game in a turn,\\
(e) find the probability that she won 2 games.
1st game 2nd game win
\hfill \mbox{\textit{Edexcel S1 2023 Q1 [11]}}