| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Two-stage sampling with replacement |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram problem with standard conditional probability calculations. Part (i) is simple justification, (ii) is routine completion, (iii) involves solving a basic linear equation from given probability, and (iv) requires Bayes' theorem application—all standard S1 techniques with no novel insight required. |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| number of balls in \(B\) is \(5 + x + 1 = x + 6\); \(P(Y) = \frac{x}{x+6}\) AG | B1 [1] | Sensible reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Box \(A\): \(\frac{8}{10}\) and \(\frac{2}{10}\) branches with labels \(W\) and \(Y\) | B1 | Both correct for box A |
| Box \(B\) from \(W\): \(\frac{6}{x+6} \to W\) and \(\frac{x+1}{x+6} \to Y\) (branch from \(W\) arm) | B1 | 1 correct |
| Box \(B\) from \(Y\): \(\frac{5}{x+6} \to W\) | B1 | 1 correct |
| \(\frac{x+1}{x+6} \to Y\) (branch from \(Y\) arm) | B1 [4] | 1 correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(W_B) = \frac{6}{x+6} = \frac{1}{3}\) | M1 | their \(\frac{6}{x+6} = \frac{1}{3}\) or \(x/x+6 = 2/3\) |
| \(x = 12\) AG | A1 [2] | Verification or solving legit |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(Y) = \frac{8}{10} \times \frac{12}{18} + \frac{2}{10} \times \frac{13}{18}\) | M1 | Attempt at \(P(Y)\) involving 2 two-factor fractions, seen anywhere |
| \(= \frac{61}{90}\) | A1 | Correct \(P(Y)\) seen as num or denom of a fraction |
| \(P(AY \mid BY) = \frac{P(AY \cap BY)}{P(Y)}\) | B1 | \(\frac{2}{10} \times \frac{13}{18}\) seen as num or denom of a fraction |
| \(= \frac{2}{10} \times \frac{13}{18} \div \frac{61}{90}\) | ||
| \(= \frac{13}{61}\ (0.213)\) | A1 [4] | Correct answer |
## Question 7:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| number of balls in $B$ is $5 + x + 1 = x + 6$; $P(Y) = \frac{x}{x+6}$ **AG** | B1 **[1]** | Sensible reason |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Box $A$: $\frac{8}{10}$ and $\frac{2}{10}$ branches with labels $W$ and $Y$ | B1 | Both correct for box A |
| Box $B$ from $W$: $\frac{6}{x+6} \to W$ and $\frac{x+1}{x+6} \to Y$ (branch from $W$ arm) | B1 | 1 correct |
| Box $B$ from $Y$: $\frac{5}{x+6} \to W$ | B1 | 1 correct |
| $\frac{x+1}{x+6} \to Y$ (branch from $Y$ arm) | B1 **[4]** | 1 correct |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(W_B) = \frac{6}{x+6} = \frac{1}{3}$ | M1 | their $\frac{6}{x+6} = \frac{1}{3}$ or $x/x+6 = 2/3$ |
| $x = 12$ **AG** | A1 **[2]** | Verification or solving legit |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(Y) = \frac{8}{10} \times \frac{12}{18} + \frac{2}{10} \times \frac{13}{18}$ | M1 | Attempt at $P(Y)$ involving 2 two-factor fractions, seen anywhere |
| $= \frac{61}{90}$ | A1 | Correct $P(Y)$ seen as num or denom of a fraction |
| $P(AY \mid BY) = \frac{P(AY \cap BY)}{P(Y)}$ | B1 | $\frac{2}{10} \times \frac{13}{18}$ seen as num or denom of a fraction |
| $= \frac{2}{10} \times \frac{13}{18} \div \frac{61}{90}$ | | |
| $= \frac{13}{61}\ (0.213)$ | A1 **[4]** | Correct answer |7 Box $A$ contains 8 white balls and 2 yellow balls. Box $B$ contains 5 white balls and $x$ yellow balls. A ball is chosen at random from box $A$ and placed in box $B$. A ball is then chosen at random from box $B$. The tree diagram below shows the possibilities for the colours of the balls chosen.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Box $A$}
\includegraphics[alt={},max width=\textwidth]{60a9d5d4-0a6a-43e2-9828-03ea2a76ed8a-3_451_874_1774_639}
\end{center}
\end{figure}
(i) Justify the probability $\frac { x } { x + 6 }$ on the tree diagram.\\
(ii) Copy and complete the tree diagram.\\
(iii) If the ball chosen from box $A$ is white then the probability that the ball chosen from box $B$ is also white is $\frac { 1 } { 3 }$. Show that the value of $x$ is 12 .\\
(iv) Given that the ball chosen from box $B$ is yellow, find the conditional probability that the ball chosen from box $A$ was yellow.
\hfill \mbox{\textit{CAIE S1 2013 Q7 [11]}}