| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Probability distribution from conditional setup |
| Difficulty | Standard +0.3 Part (a) is trivial algebra using sum of probabilities equals 1. Part (b) requires systematic enumeration of cases where three scores sum to ≤4, which is straightforward but requires care. Part (c) involves conditional probability with enumeration of products ≤4 given X is odd—more steps but still routine application of P(A|B) formula with no novel insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.04a Discrete probability distributions |
| \(x\) | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | 0.28 | \(p\) | \(2 p\) | \(3 p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.28 + 6p = 1,\ p = 0.12\) | B1 | Using sum of probabilities \(= 1\). Accept \(0.28 + p + 2p + 3p = 1\), \(p = 0.12\). Substitution of \(0.12\) into expression scores B0 |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For fair spinners (blue and green), probability of any score is \(0.25\). Scenarios to give total 4 or less: | B1 | Correct probability for 1 identified scenario, accept unsimplified |
| \(R=1, B=1, G=2\): \(0.28\times(0.25)^2 = 0.0175\) | M1 | Add values of 4 correct scenarios, may be implied by correct unsimplified expressions. No incorrect/repeated scenarios |
| \(R=1, B=2, G=1\): \(0.28\times(0.25)^2 = 0.0175\) | ||
| \(R=1, B=1, G=1\): \(0.28\times(0.25)^2 = 0.0175\) | ||
| \(R=2, B=1, G=1\): \(0.12\times(0.25)^2 = 0.0075\) | ||
| \(= 0.06\) | A1 | If A0 scored, SC B1 for \(0.06\) www |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X \text{ is odd}) = 0.28 + 2\times0.12\ \text{or}\ 0.24 = 0.52\) | B1 | Seen alone or as denominator of conditional probability fraction. Accept unsimplified |
| Scenarios listed (R=1,B=1,G=1 through 4; R=1,B=2,G=1 and 2; R=1,B=3,G=1; R=1,B=4,G=1; R=3,B=1,G=1), each \(0.28\times(0.25)^2 = 0.0175\) or \(0.24\times(0.25)^2 = 0.015\) | M1 | Values of at least 5 identified correct scenarios added, accept unsimplified, condone incorrect scenarios in calculation |
| \(P(\text{product} \leqslant 4 \cap X \text{ is odd}) = 0.28\times(0.25)^2\times8 + 0.24\times(0.25)^2\) | M1 | \(0.28\times(0.25)^2\times x + 0.24\times(0.25)^2\), or \(0.0175\times x + 0.015\) where \(x=4,5,6,7,\) or \(8\). Seen alone or as numerator/denominator of conditional probability fraction |
| \(P(\text{product} \leqslant 4 \mid X \text{ is odd}) = \frac{P(\text{product} \leqslant 4 \cap X \text{ is odd})}{P(X \text{ is odd})} = \frac{0.155}{0.52}\) | M1 | \(\frac{0.28\times(0.25)^2\times x + 0.24\times(0.25)^2}{0.28+0.24}\), \(x=4,5,6,7,8\) |
| \(= 0.298,\ \frac{155}{520},\ \frac{31}{104}\) | A1 | \(0.2980769\ldots\) to at least 3SF |
| Total: 5 marks |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.28 + 6p = 1,\ p = 0.12$ | **B1** | Using sum of probabilities $= 1$. Accept $0.28 + p + 2p + 3p = 1$, $p = 0.12$. Substitution of $0.12$ into expression scores B0 |
| **Total: 1 mark** | | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For fair spinners (blue and green), probability of any score is $0.25$. Scenarios to give total 4 or less: | **B1** | Correct probability for 1 identified scenario, accept unsimplified |
| $R=1, B=1, G=2$: $0.28\times(0.25)^2 = 0.0175$ | **M1** | Add values of 4 correct scenarios, may be implied by correct unsimplified expressions. No incorrect/repeated scenarios |
| $R=1, B=2, G=1$: $0.28\times(0.25)^2 = 0.0175$ | | |
| $R=1, B=1, G=1$: $0.28\times(0.25)^2 = 0.0175$ | | |
| $R=2, B=1, G=1$: $0.12\times(0.25)^2 = 0.0075$ | | |
| $= 0.06$ | **A1** | If A0 scored, **SC B1** for $0.06$ www |
| **Total: 3 marks** | | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \text{ is odd}) = 0.28 + 2\times0.12\ \text{or}\ 0.24 = 0.52$ | **B1** | Seen alone or as denominator of conditional probability fraction. Accept unsimplified |
| Scenarios listed (R=1,B=1,G=1 through 4; R=1,B=2,G=1 and 2; R=1,B=3,G=1; R=1,B=4,G=1; R=3,B=1,G=1), each $0.28\times(0.25)^2 = 0.0175$ or $0.24\times(0.25)^2 = 0.015$ | **M1** | Values of at least 5 identified correct scenarios added, accept unsimplified, condone incorrect scenarios in calculation |
| $P(\text{product} \leqslant 4 \cap X \text{ is odd}) = 0.28\times(0.25)^2\times8 + 0.24\times(0.25)^2$ | **M1** | $0.28\times(0.25)^2\times x + 0.24\times(0.25)^2$, or $0.0175\times x + 0.015$ where $x=4,5,6,7,$ or $8$. Seen alone or as numerator/denominator of conditional probability fraction |
| $P(\text{product} \leqslant 4 \mid X \text{ is odd}) = \frac{P(\text{product} \leqslant 4 \cap X \text{ is odd})}{P(X \text{ is odd})} = \frac{0.155}{0.52}$ | **M1** | $\frac{0.28\times(0.25)^2\times x + 0.24\times(0.25)^2}{0.28+0.24}$, $x=4,5,6,7,8$ |
| $= 0.298,\ \frac{155}{520},\ \frac{31}{104}$ | **A1** | $0.2980769\ldots$ to at least 3SF |
| **Total: 5 marks** | | |
---5 A red spinner has four sides labelled $1,2,3,4$. When the spinner is spun, the score is the number on the side on which it lands. The random variable $X$ denotes this score. The probability distribution table for $X$ is given below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & 0.28 & $p$ & $2 p$ & $3 p$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $p = 0.12$.\\
A fair blue spinner and a fair green spinner each have four sides labelled 1, 2, 3, 4. All three spinners (red, blue and green) are spun at the same time.
\item Find the probability that the sum of the three scores is 4 or less.
\item Find the probability that the product of the three scores is 4 or less given that $X$ is odd.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q5 [9]}}