| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Dice/random device selects population |
| Difficulty | Standard +0.3 This is a standard conditional probability question with clear structure: part (a) requires law of total probability with straightforward counting of combinations from two bags, and part (b) applies Bayes' theorem. The calculations involve basic combinations and probability rules taught in S1, with no conceptual surprises or complex multi-step reasoning beyond the standard template for such problems. |
| 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 |
| \([P(WW) = P(AWW) + P(BWW) =]\) \(\frac{2}{6}\times\frac{8}{15}\times\frac{7}{14} + \frac{4}{6}\times\frac{6}{15}\times\frac{5}{14}\) | M1 | Either \(\frac{2}{6}\times\frac{8}{15}\times\frac{7}{14}\) or \(\frac{4}{6}\times\frac{6}{15}\times\frac{5}{14}\) seen, accept unsimplified |
| M1 | \(\frac{q}{6}\times\frac{r}{15}\times\frac{r-1}{14} + \frac{6-q}{6}\times\frac{s}{15}\times\frac{s-1}{14}\) seen, no additional terms, accept unsimplified. Condone \(\frac{q}{6}\times\frac{r}{15}\times\frac{r}{15} + \frac{6-q}{6}\times\frac{s}{15}\times\frac{s}{15}\), \(1 \leq q \leq 5\), \(1 < r, s < 9\) | |
| \(\left[= \frac{56}{630} + \frac{60}{630} = \frac{4}{45} + \frac{2}{21}\right] = \frac{58}{315}\) or \(0.184\) | A1 | SC B1 for \(58/315\) if either M mark withheld |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(B \mid WR \text{ or } RW) = \frac{P(W \text{ & } R \text{ from bag B})}{P(W \text{ and } R)}\) | B1 | \(P(W\) & \(R\) from bag \(B) = \frac{2}{3}\times\frac{6}{15}\times\frac{7}{14} + \frac{2}{3}\times\frac{7}{15}\times\frac{6}{14}\) or \(2\times\frac{2}{3}\times\frac{6}{15}\times\frac{7}{14}\) \(\left[= \frac{4}{15} \text{ or } 0.267\right]\). Seen alone or as numerator/denominator of conditional probability |
| \(\dfrac{\frac{4}{6}\times\frac{6}{15}\times\frac{7}{14} + \frac{4}{6}\times\frac{7}{15}\times\frac{6}{14}}{\frac{2}{6}\times\frac{8}{15}\times\frac{4}{14}+\frac{2}{6}\times\frac{4}{15}\times\frac{8}{14}+\frac{4}{6}\times\frac{6}{15}\times\frac{7}{14}+\frac{4}{6}\times\frac{7}{15}\times\frac{6}{14}}\) | M1 | \(P(WR \text{ or } RW) = P(W\) & \(R\) from bag \(A) + P(W\) & \(R\) from bag \(B)\) \(= a\times\frac{2}{6}\times\frac{8}{15}\times\frac{4}{14} + a\times\frac{4}{6}\times\frac{6}{15}\times\frac{7}{14}\) or \(= a\times\frac{2}{6}\times\frac{8}{15}\times\frac{4}{14} + \textit{their } P(W\) & \(R\) from bag \(B)\). \(a = 1\) or \(2\). [expect \(\frac{116}{315}\) or \(0.368\)]. Seen alone or as numerator/denominator of conditional probability |
| \(\dfrac{168/630}{232/630} = \dfrac{4/15}{116/315}\) | M1 | \(\frac{\textit{their identified } P(W \text{ \& } R \text{ from bag } B)}{\textit{their identified } P(WR \text{ or } RW)}\). Accept unsimplified |
| \(= \frac{168}{232}, \frac{21}{29}\) or \(0.724\) | A1 | \(0.7241379\) to at least 3SF |
| 4 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(WW) = P(AWW) + P(BWW) =]$ $\frac{2}{6}\times\frac{8}{15}\times\frac{7}{14} + \frac{4}{6}\times\frac{6}{15}\times\frac{5}{14}$ | M1 | Either $\frac{2}{6}\times\frac{8}{15}\times\frac{7}{14}$ or $\frac{4}{6}\times\frac{6}{15}\times\frac{5}{14}$ seen, accept unsimplified |
| | M1 | $\frac{q}{6}\times\frac{r}{15}\times\frac{r-1}{14} + \frac{6-q}{6}\times\frac{s}{15}\times\frac{s-1}{14}$ seen, no additional terms, accept unsimplified. Condone $\frac{q}{6}\times\frac{r}{15}\times\frac{r}{15} + \frac{6-q}{6}\times\frac{s}{15}\times\frac{s}{15}$, $1 \leq q \leq 5$, $1 < r, s < 9$ |
| $\left[= \frac{56}{630} + \frac{60}{630} = \frac{4}{45} + \frac{2}{21}\right] = \frac{58}{315}$ or $0.184$ | A1 | **SC B1** for $58/315$ if either M mark withheld |
| | **3** | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(B \mid WR \text{ or } RW) = \frac{P(W \text{ & } R \text{ from bag B})}{P(W \text{ and } R)}$ | B1 | $P(W$ & $R$ from bag $B) = \frac{2}{3}\times\frac{6}{15}\times\frac{7}{14} + \frac{2}{3}\times\frac{7}{15}\times\frac{6}{14}$ or $2\times\frac{2}{3}\times\frac{6}{15}\times\frac{7}{14}$ $\left[= \frac{4}{15} \text{ or } 0.267\right]$. Seen alone or as numerator/denominator of conditional probability |
| $\dfrac{\frac{4}{6}\times\frac{6}{15}\times\frac{7}{14} + \frac{4}{6}\times\frac{7}{15}\times\frac{6}{14}}{\frac{2}{6}\times\frac{8}{15}\times\frac{4}{14}+\frac{2}{6}\times\frac{4}{15}\times\frac{8}{14}+\frac{4}{6}\times\frac{6}{15}\times\frac{7}{14}+\frac{4}{6}\times\frac{7}{15}\times\frac{6}{14}}$ | M1 | $P(WR \text{ or } RW) = P(W$ & $R$ from bag $A) + P(W$ & $R$ from bag $B)$ $= a\times\frac{2}{6}\times\frac{8}{15}\times\frac{4}{14} + a\times\frac{4}{6}\times\frac{6}{15}\times\frac{7}{14}$ or $= a\times\frac{2}{6}\times\frac{8}{15}\times\frac{4}{14} + \textit{their } P(W$ & $R$ from bag $B)$. $a = 1$ or $2$. [expect $\frac{116}{315}$ or $0.368$]. Seen alone or as numerator/denominator of conditional probability |
| $\dfrac{168/630}{232/630} = \dfrac{4/15}{116/315}$ | M1 | $\frac{\textit{their identified } P(W \text{ \& } R \text{ from bag } B)}{\textit{their identified } P(WR \text{ or } RW)}$. Accept unsimplified |
| $= \frac{168}{232}, \frac{21}{29}$ or $0.724$ | A1 | $0.7241379$ to at least 3SF |
| | **4** | |3 Tim has two bags of marbles, $A$ and $B$.\\
Bag $A$ contains 8 white, 4 red and 3 yellow marbles.\\
Bag $B$ contains 6 white, 7 red and 2 yellow marbles.\\
Tim also has an ordinary fair 6 -sided dice. He rolls the dice. If he obtains a 1 or 2 , he chooses two marbles at random from bag $A$, without replacement. If he obtains a $3,4,5$ or 6 , he chooses two marbles at random from bag $B$, without replacement.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that both marbles are white.
\item Find the probability that the two marbles come from bag $B$ given that one is white and one is red. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q3 [7]}}