| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with given score/outcome |
| Difficulty | Standard +0.3 This is a standard two-stage conditional probability problem requiring a probability tree, the law of total probability for part (a), and Bayes' theorem for part (b). The calculations involve straightforward combinations and fractions with no conceptual surprises, making it slightly easier than average for A-level. |
| 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 | Marks | Guidance |
| \(P(HR) + P(TR) + P(TBR)\): \(\frac{1}{3} \times \frac{4}{9} + \frac{2}{3} \times \frac{4}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}\) | B1 | Two of the calculations for \(P(HR)\), \(P(TBR)\), either \(P(TR)\) or \(P(TRR) + P(TRB)\) unsimplified. Condone \(\frac{4}{8} = \frac{1}{2}\) in unsimplified calculation. Condone use of tree diagram if values correct at end |
| \(P(HR) + (P(TRR) + P(TRB)) + P(TBR)\): \(\frac{1}{3} \times \frac{4}{9} + \left(\frac{2}{3} \times \frac{4}{9} \times \frac{3}{8} + \frac{2}{3} \times \frac{4}{9} \times \frac{4}{8}\right) + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}\) | M1 | Values of all correct identified scenarios added. Correct branches may be identified on the tree diagram |
| \(\left[= \frac{4}{27} + \frac{8}{27} + \frac{5}{27}\right] = \frac{17}{27}\) | A1 | \(0.6296\ldots, 0.630\). If M0 scored SC B1 for acceptable answers, WWW |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - P(HB) - P(TBB) = 1 - \left(\frac{1}{3} \times \frac{5}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}\right) = \left[1 - \frac{5}{27} - \frac{5}{27}\right]\) | (B1) | One calculation of \(P(HB)\), \(P(TBB)\) unsimplified. \(1 -\) probability must be seen. Condone use of tree diagram if values correct at end |
| (M1) | \(1 -\) values of two correct identified scenarios subtracted. Correct branches may be identified on the tree diagram | |
| \(= \frac{17}{27}\) | (A1) | \(0.6296\ldots, 0.630\). If M0 scored SC B1 for acceptable answers, WWW |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(HR) + P(T, (1 - \text{no } R)) = \frac{1}{3} \times \frac{4}{9} + \frac{2}{3}\left(1 - \left(\frac{5}{9} \times \frac{4}{8}\right)\right)\) | (B1) | Calculation for \(P(T, (1 - \text{no } R))\) seen unsimplified. Condone use of tree diagram if values correct at end |
| \(\left[\frac{4}{27} + \frac{2}{3}\left(1 - \frac{20}{27}\right)\right]\) | (M1) | Values of two correct identified scenarios added. Correct branches may be identified on the tree diagram |
| \(= \frac{17}{27}\) | (A1) | \(0.6296\ldots, 0.630\). If M0 scored SC B1 for acceptable answers, WWW |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{head} \mid \text{no reds}) = \dfrac{P(\text{head} \cap \text{no reds})}{P(\text{no reds})} = \dfrac{\frac{1}{3} \times \frac{5}{9}}{1 - \frac{17}{27}} = \dfrac{\frac{5}{27}}{\frac{10}{27}}\) | M1 | \(\dfrac{d}{1-\text{their}(\mathbf{a})}\) or \(\dfrac{d}{1-\frac{17}{27}}\) or \(\dfrac{d}{\frac{10}{27}}\), \(0 < d < 1\). Condone \(\frac{10}{27} = 0.3704\) or more accurate. |
| \(= \dfrac{1}{2}\) | A1 | OE. Condone \(0.499[9\ldots]\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{head} \mid \text{no reds}) = \dfrac{P(\text{head} \cap \text{blue})}{P(\text{HB})+P(\text{TBB})} = \dfrac{\frac{1}{3} \times \frac{5}{9}}{\frac{1}{3} \times \frac{5}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}} = \dfrac{\frac{5}{27}}{\frac{10}{27}}\) | (M1) | \(\dfrac{d}{\frac{1}{3} \times \frac{5}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}}\) or \(\dfrac{d}{\frac{10}{27}}\), \(0 < d < 1\). Condone \(\frac{10}{27} = 0.3704\) or more accurate. |
| \(= \dfrac{1}{2}\) | (A1) | OE. Condone \(0.499[9\ldots]\). |
## Question 2:
**Part (a) — Method 1**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(HR) + P(TR) + P(TBR)$: $\frac{1}{3} \times \frac{4}{9} + \frac{2}{3} \times \frac{4}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}$ | B1 | Two of the calculations for $P(HR)$, $P(TBR)$, either $P(TR)$ or $P(TRR) + P(TRB)$ unsimplified. Condone $\frac{4}{8} = \frac{1}{2}$ in unsimplified calculation. Condone use of tree diagram if values correct at end |
| $P(HR) + (P(TRR) + P(TRB)) + P(TBR)$: $\frac{1}{3} \times \frac{4}{9} + \left(\frac{2}{3} \times \frac{4}{9} \times \frac{3}{8} + \frac{2}{3} \times \frac{4}{9} \times \frac{4}{8}\right) + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}$ | M1 | Values of all correct identified scenarios added. Correct branches may be identified on the tree diagram |
| $\left[= \frac{4}{27} + \frac{8}{27} + \frac{5}{27}\right] = \frac{17}{27}$ | A1 | $0.6296\ldots, 0.630$. If M0 scored **SC B1** for acceptable answers, WWW |
| | **Total: 3** | |
**Part (a) — Method 2**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - P(HB) - P(TBB) = 1 - \left(\frac{1}{3} \times \frac{5}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}\right) = \left[1 - \frac{5}{27} - \frac{5}{27}\right]$ | (B1) | One calculation of $P(HB)$, $P(TBB)$ unsimplified. $1 -$ probability must be seen. Condone use of tree diagram if values correct at end |
| | (M1) | $1 -$ values of two correct identified scenarios subtracted. Correct branches may be identified on the tree diagram |
| $= \frac{17}{27}$ | (A1) | $0.6296\ldots, 0.630$. If M0 scored **SC B1** for acceptable answers, WWW |
| | **Total: 3** | |
**Part (a) — Method 3**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(HR) + P(T, (1 - \text{no } R)) = \frac{1}{3} \times \frac{4}{9} + \frac{2}{3}\left(1 - \left(\frac{5}{9} \times \frac{4}{8}\right)\right)$ | (B1) | Calculation for $P(T, (1 - \text{no } R))$ seen unsimplified. Condone use of tree diagram if values correct at end |
| $\left[\frac{4}{27} + \frac{2}{3}\left(1 - \frac{20}{27}\right)\right]$ | (M1) | Values of two correct identified scenarios added. Correct branches may be identified on the tree diagram |
| $= \frac{17}{27}$ | (A1) | $0.6296\ldots, 0.630$. If M0 scored **SC B1** for acceptable answers, WWW |
| | **Total: 3** | |
## Question 2(b):
**Method 1:**
$P(\text{head} \mid \text{no reds}) = \dfrac{P(\text{head} \cap \text{no reds})}{P(\text{no reds})} = \dfrac{\frac{1}{3} \times \frac{5}{9}}{1 - \frac{17}{27}} = \dfrac{\frac{5}{27}}{\frac{10}{27}}$ | M1 | $\dfrac{d}{1-\text{their}(\mathbf{a})}$ or $\dfrac{d}{1-\frac{17}{27}}$ or $\dfrac{d}{\frac{10}{27}}$, $0 < d < 1$. Condone $\frac{10}{27} = 0.3704$ or more accurate.
$= \dfrac{1}{2}$ | A1 | OE. Condone $0.499[9\ldots]$.
**Method 2:**
$P(\text{head} \mid \text{no reds}) = \dfrac{P(\text{head} \cap \text{blue})}{P(\text{HB})+P(\text{TBB})} = \dfrac{\frac{1}{3} \times \frac{5}{9}}{\frac{1}{3} \times \frac{5}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}} = \dfrac{\frac{5}{27}}{\frac{10}{27}}$ | (M1) | $\dfrac{d}{\frac{1}{3} \times \frac{5}{9} + \frac{2}{3} \times \frac{5}{9} \times \frac{4}{8}}$ or $\dfrac{d}{\frac{10}{27}}$, $0 < d < 1$. Condone $\frac{10}{27} = 0.3704$ or more accurate.
$= \dfrac{1}{2}$ | (A1) | OE. Condone $0.499[9\ldots]$.
**Total: 2 marks**
---2 Seva has a coin which is biased so that when it is thrown the probability of obtaining a head is $\frac { 1 } { 3 }$. He also has a bag containing 4 red marbles and 5 blue marbles.
Seva throws the coin. If he obtains a head, he selects one marble from the bag at random. If he obtains a tail, he selects two marbles from the bag at random and without replacement.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Seva selects at least one red marble.
\item Find the probability that Seva obtains a head given that he selects no red marbles.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q2 [5]}}