| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Sequential selection without replacement |
| Difficulty | Standard +0.3 This is a standard tree diagram problem with sequential selection without replacement, requiring careful tracking of changing probabilities after the first draw and the addition of a red marble. Part (c) involves conditional probability using Bayes' theorem. While it requires multiple steps and careful bookkeeping, it follows a well-established template for this topic with no novel insights needed—slightly easier than average due to its routine nature. |
| Spec | 2.03b 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 1st column: 2 branches \(X\), \(Y\) with probabilities \(\frac{1}{2}\), \(\frac{1}{2}\) | B1 | 1st column, 2 branches identified \(X\), \(Y\) with probabilities \(\frac{1}{2}\), \(\frac{1}{2}\) indicated |
| 2nd column: 4 branches identified \(R\ B\ R\ B\) (oe) with probabilities \(\frac{7}{10}, \frac{3}{10}, \frac{4}{5}, \frac{1}{5}\) indicated appropriately | B1 | 2nd column (1st marble pick) of 4 branches identified \(R\ B\ R\ B\) (oe) and probabilities \(\frac{7}{10}, \frac{3}{10}, \frac{4}{5}, \frac{1}{5}\) indicated appropriately |
| 3rd column: 8 branches identified \(R\ B\ R\ B\ R\ B\ R\ [B]\) (oe) with probabilities \(\frac{7}{10}, \frac{3}{10}, \frac{8}{10}, \frac{2}{10}, \frac{4}{5}, \frac{1}{5}, 1, [0]\) | B1 | 3rd column (2nd marble pick) of 8 branches identified; condone omission of \(YBB\) branch if \(YBR\) branch fully correct; ignore any additional columns; if separate tree diagrams for bags \(X\) and \(Y\), B0B1B1 max if bags clearly identified |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}\times\frac{3}{10}\times\frac{2}{10} + \frac{1}{2}\times\frac{7}{10}\times\frac{7}{10} + \frac{1}{2}\times\frac{4}{5}\times\frac{4}{5}\) | B1 FT | \(\left[P(BB)=\right]\frac{1}{2}\times\frac{3}{10}\times\frac{2}{10}\left[+\frac{1}{2}\times\frac{1}{5}\times 0\right] = \frac{6}{200}\) seen; accept unsimplified; FT from 6(a) unsimplified only with 3 term probabilities |
| \(\left[= \frac{6}{200} + \frac{49}{200} + \frac{16}{50},\ 0.03 + 0.245 + 0.32\right]\) | B1 FT | Either \([P(XRR) =] \frac{1}{2}\times\frac{7}{10}\times\frac{7}{10}\) or \([P(YRR) =] \frac{1}{2}\times\frac{4}{5}\times\frac{4}{5}\) seen; FT from 6(a) unsimplified only with 3 term probabilities |
| \([P(BB) + P(XRR) + P(YRR) =]\) their \(\frac{6}{200}\) + their \(\frac{49}{200}\) + their \(\frac{16}{50}\) | M1 | Accept unsimplified, consistent with tree diagram if not clearly identified by notation |
| \(= \frac{119}{200},\ 0.595\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\frac{1}{2}\times\frac{4}{5}\times\frac{1}{5}+\frac{1}{2}\times\frac{1}{5}[\times 1]}{1-\text{their}\left(\frac{119}{200}\right)}\) or \(\frac{\frac{1}{2}\times\frac{4}{5}\times\frac{1}{5}+\frac{1}{2}\times\frac{1}{5}[\times 1]}{\frac{1}{2}\times\frac{7}{10}\times\frac{3}{10}+\frac{1}{2}\times\frac{3}{10}\times\frac{8}{10}+\frac{1}{2}\times\frac{4}{5}\times\frac{1}{5}+\frac{1}{2}\times\frac{1}{5}[\times 1]}\) | M1 | FT from their 6(a) and their 6(b) with 3 term probabilities unsimplified only or correct; accept \(\frac{4}{50}+\frac{1}{10},\ \frac{2}{25}+\frac{1}{10},\ 0.08+0.1\) |
| \(= \left[\frac{\frac{9}{50}}{\frac{81}{200}}\right] = \frac{4}{9},\ 0.444\) | A1 | Accept \(\frac{36}{81}\), \(0.\dot{4}\) |
## Question 6(a):
Tree diagram with 1st column: 2 branches $X$, $Y$ with probabilities $\frac{1}{2}$, $\frac{1}{2}$ | B1 | 1st column, 2 branches identified $X$, $Y$ with probabilities $\frac{1}{2}$, $\frac{1}{2}$ indicated
2nd column: 4 branches identified $R\ B\ R\ B$ (oe) with probabilities $\frac{7}{10}, \frac{3}{10}, \frac{4}{5}, \frac{1}{5}$ indicated appropriately | B1 | 2nd column (1st marble pick) of 4 branches identified $R\ B\ R\ B$ (oe) and probabilities $\frac{7}{10}, \frac{3}{10}, \frac{4}{5}, \frac{1}{5}$ indicated appropriately
3rd column: 8 branches identified $R\ B\ R\ B\ R\ B\ R\ [B]$ (oe) with probabilities $\frac{7}{10}, \frac{3}{10}, \frac{8}{10}, \frac{2}{10}, \frac{4}{5}, \frac{1}{5}, 1, [0]$ | B1 | 3rd column (2nd marble pick) of 8 branches identified; condone omission of $YBB$ branch if $YBR$ branch fully correct; ignore any additional columns; if separate tree diagrams for bags $X$ and $Y$, B0B1B1 max if bags clearly identified
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## Question 6(b):
$[P(\text{both same colour}) = P(BB) + P(RR) = P(XBB) + P(XRR) + P(YRR) =]$
$\frac{1}{2}\times\frac{3}{10}\times\frac{2}{10} + \frac{1}{2}\times\frac{7}{10}\times\frac{7}{10} + \frac{1}{2}\times\frac{4}{5}\times\frac{4}{5}$ | B1 FT | $\left[P(BB)=\right]\frac{1}{2}\times\frac{3}{10}\times\frac{2}{10}\left[+\frac{1}{2}\times\frac{1}{5}\times 0\right] = \frac{6}{200}$ seen; accept unsimplified; FT from 6(a) unsimplified only with 3 term probabilities
$\left[= \frac{6}{200} + \frac{49}{200} + \frac{16}{50},\ 0.03 + 0.245 + 0.32\right]$ | B1 FT | Either $[P(XRR) =] \frac{1}{2}\times\frac{7}{10}\times\frac{7}{10}$ or $[P(YRR) =] \frac{1}{2}\times\frac{4}{5}\times\frac{4}{5}$ seen; FT from 6(a) unsimplified only with 3 term probabilities
$[P(BB) + P(XRR) + P(YRR) =]$ their $\frac{6}{200}$ + their $\frac{49}{200}$ + their $\frac{16}{50}$ | M1 | Accept unsimplified, consistent with tree diagram if not clearly identified by notation
$= \frac{119}{200},\ 0.595$ | A1 |
**Special case:** if $\frac{1}{2}$ omitted consistently in tree diagram and calculation, no FT: SC B1 $[P(BB) =] \frac{3}{10}\times\frac{2}{10}\left[\frac{1}{5}\times 0\right]$; SC B1 $[P(RR) =] \frac{7}{10}\times\frac{7}{10}+\frac{4}{5}\times\frac{4}{5}$; SC B1 $\frac{3}{10}\times\frac{2}{10}\left[+\frac{1}{5}\times 0\right]+\frac{7}{10}\times\frac{7}{10}+\frac{4}{5}\times\frac{4}{5}$
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## Question 6(c):
$\left[P(\text{bag }Y \mid \text{different colours}) = \left(\frac{P(\text{bag }Y \cap \text{different colours})}{P(\text{different colours})}\right)\right]$
$\frac{\frac{1}{2}\times\frac{4}{5}\times\frac{1}{5}+\frac{1}{2}\times\frac{1}{5}[\times 1]}{1-\text{their}\left(\frac{119}{200}\right)}$ or $\frac{\frac{1}{2}\times\frac{4}{5}\times\frac{1}{5}+\frac{1}{2}\times\frac{1}{5}[\times 1]}{\frac{1}{2}\times\frac{7}{10}\times\frac{3}{10}+\frac{1}{2}\times\frac{3}{10}\times\frac{8}{10}+\frac{1}{2}\times\frac{4}{5}\times\frac{1}{5}+\frac{1}{2}\times\frac{1}{5}[\times 1]}$ | M1 | FT from their 6(a) and their 6(b) with 3 term probabilities unsimplified only or correct; accept $\frac{4}{50}+\frac{1}{10},\ \frac{2}{25}+\frac{1}{10},\ 0.08+0.1$
$= \left[\frac{\frac{9}{50}}{\frac{81}{200}}\right] = \frac{4}{9},\ 0.444$ | A1 | Accept $\frac{36}{81}$, $0.\dot{4}$
**Special case:** if $\frac{1}{2}$ omitted consistently, no FT: SC B1 $\frac{\frac{4}{5}\times\frac{1}{5}+\frac{1}{5}[\times 1]}{1-\text{their \textbf{6(b)}}}$ or $\frac{\frac{4}{5}\times\frac{1}{5}+\frac{1}{5}[\times 1]}{\frac{7}{10}\times\frac{3}{10}+\frac{3}{10}\times\frac{8}{10}+\frac{4}{5}\times\frac{1}{5}+\frac{1}{5}[\times 1]}$6 Freddie has two bags of marbles.\\
Bag $X$ contains 7 red marbles and 3 blue marbles.\\
Bag $Y$ contains 4 red marbles and 1 blue marble.\\
Freddie chooses one of the bags at random. A marble is removed at random from that bag and not replaced. A new red marble is now added to each bag. A second marble is then removed at random from the same bag that the first marble had been removed from.
\begin{enumerate}[label=(\alph*)]
\item Draw a tree diagram to represent this information, showing the probability on each of the branches.
\item Find the probability that both of the marbles removed from the bag are the same colour.
\item Find the probability that bag $Y$ is chosen given that the marbles removed are not both the same colour.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q6 [9]}}