| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Transfer between containers |
| Difficulty | Moderate -0.3 This is a standard S1 tree diagram question with sequential probability calculations. While it requires careful tracking through multiple stages and conditional probability in part (c), the techniques are routine for this module. The calculations are straightforward once the tree is set up, and part (b) provides scaffolding for the probability distribution in part (d). Slightly easier than average due to the structured guidance through parts (a)-(b). |
| 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 principles5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram: branches 2, 4 then 7 or 8 with labels \(Y\) or \(R\) | B1 | Completing structure of branches |
| Correct probabilities for at least bag A and bag B | B1 | Allow exact decimals |
| Fully correct tree diagram | B1 | Condone missing 0 as probability |
| Answer | Marks | Guidance |
|---|---|---|
| Cases \(RYY\) or \(YRY\) or \(YYR\): \(\text{Prob} = \frac{2}{3} \times \frac{1}{5} \times \frac{1}{8} + \frac{1}{3} \times \frac{4}{5} \times \frac{1}{8} + \frac{1}{3} \times \frac{1}{5} \times \frac{6}{8}\) | M1 A1ft | At least one correct product of 3 probabilities; all 3 products added (no extras) |
| \(= \frac{1}{120}(2 + 4 + 6) = \frac{12}{120} = \frac{1}{10}\) | A1*cso | Fully correct solution with no incorrect statements |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(RYY \mid RYY \text{ or } YRY \text{ or } YYR) = \dfrac{\frac{2}{3} \times \frac{1}{5} \times \frac{1}{8}}{\frac{1}{10}}\) | M1 | Ratio of probabilities with denominator 0.1 and numerator \(\frac{1}{60}\) |
| \(= \dfrac{1}{6}\) | A1 | For \(\frac{1}{6}\) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\): 0, 1, 2, 3 | B1 | Correct sample space \(\{0,1,2,3\}\) |
| \(P(X=x)\): \(\frac{64}{120}\) or \(\frac{8}{15}\), \(\frac{42}{120}\) or \(\frac{7}{20}\), \(\left[\frac{1}{10}\right]\), \(\frac{2}{120}\) or \(\frac{1}{60}\) | M1, A1 | At least 1 correct value and associated probability; fully correct distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \frac{1}{120}(0 + \text{"42"} + \text{"2} \times 12\text{"} + \text{"3} \times 2\text{"})\) | M1 | Attempt at correct expression (at least 2 correct ft part (d) non-zero products) |
| \(= \frac{72}{120}\) or 0.6 | A1 | For 0.6 or any exact equivalent |
# Question 4:
## Part (a)
| Tree diagram: branches 2, 4 then 7 or 8 with labels $Y$ or $R$ | B1 | Completing structure of branches |
| Correct probabilities for at least bag **A** and bag **B** | B1 | Allow exact decimals |
| Fully correct tree diagram | B1 | Condone missing 0 as probability |
## Part (b)
| Cases $RYY$ or $YRY$ or $YYR$: $\text{Prob} = \frac{2}{3} \times \frac{1}{5} \times \frac{1}{8} + \frac{1}{3} \times \frac{4}{5} \times \frac{1}{8} + \frac{1}{3} \times \frac{1}{5} \times \frac{6}{8}$ | M1 A1ft | At least one correct product of 3 probabilities; all 3 products added (no extras) |
| $= \frac{1}{120}(2 + 4 + 6) = \frac{12}{120} = \frac{1}{10}$ | A1*cso | Fully correct solution with no incorrect statements |
## Part (c)
| $P(RYY \mid RYY \text{ or } YRY \text{ or } YYR) = \dfrac{\frac{2}{3} \times \frac{1}{5} \times \frac{1}{8}}{\frac{1}{10}}$ | M1 | Ratio of probabilities with denominator 0.1 and numerator $\frac{1}{60}$ |
| $= \dfrac{1}{6}$ | A1 | For $\frac{1}{6}$ or exact equivalent |
## Part (d)
| $x$: 0, 1, 2, 3 | B1 | Correct sample space $\{0,1,2,3\}$ |
| $P(X=x)$: $\frac{64}{120}$ or $\frac{8}{15}$, $\frac{42}{120}$ or $\frac{7}{20}$, $\left[\frac{1}{10}\right]$, $\frac{2}{120}$ or $\frac{1}{60}$ | M1, A1 | At least 1 correct value and associated probability; fully correct distribution |
## Part (e)
| $E(X) = \frac{1}{120}(0 + \text{"42"} + \text{"2} \times 12\text{"} + \text{"3} \times 2\text{"})$ | M1 | Attempt at correct expression (at least 2 correct ft part (d) non-zero products) |
| $= \frac{72}{120}$ or **0.6** | A1 | For 0.6 or any exact equivalent |
---4. Three bags A, B and $\mathbf { C }$ each contain coloured balls.
Bag A contains 4 red balls and 2 yellow balls only.\\
Bag B contains 4 red balls and 1 yellow ball only.\\
Bag $\mathbf { C }$ contains 6 red balls only.
In a game\\
Mike takes a ball at random from bag $\mathbf { A }$, records the colour and places it in bag $\mathbf { C }$. He then takes a ball at random from bag $\mathbf { B }$, records the colour and places it in bag $\mathbf { C }$. Finally, Mike takes a ball at random from bag $\mathbf { C }$ and records the colour.
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram on the page opposite, to illustrate the game by adding the remaining branches and all probabilities.
\item Show that the probability that Mike records a yellow ball exactly twice is $\frac { 1 } { 10 }$
Given that Mike records exactly 2 yellow balls,
\item find the probability that the ball drawn from bag $\mathbf { A }$ is red.
Mike plays this game a large number of times, each time starting with the bags containing balls as described above. The random variable $X$ represents the number of yellow balls recorded in a single game.
\item Find the probability distribution of $X$
\item Find $\mathrm { E } ( X )$
Bag B\\
Bag C
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Bag A}
\includegraphics[alt={},max width=\textwidth]{29ac0c0b-f963-40a1-beba-7146bbb2d021-13_739_1580_411_182}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2021 Q4 [13]}}