| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring systematic enumeration of outcomes for three biased coins. Part (a) guides students through one calculation, part (b) requires completing similar calculations for all cases (0-3 heads), and part (c) is a standard variance calculation using the given expectation. While it involves multiple steps and careful arithmetic with fractions, it follows a completely standard template with no novel insight required, making it slightly easier than average. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| HHT: \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{5} = \frac{4}{45}\) | M1 | One 3 factor probability with 3, 3, 5 as denominators |
| HTH: \(\frac{2}{3} \times \frac{1}{3} \times \frac{4}{5} = \frac{8}{45}\) | M1 | 3 factor probabilities for 2 or 3 correct scenarios added, no incorrect scenarios |
| THH: \(\frac{1}{3} \times \frac{2}{3} \times \frac{4}{5} = \frac{8}{45}\) | ||
| Total \(= \frac{20}{45} = \frac{4}{9}\) | A1 | AG, Total of 3 products with clear context |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with \(x\): 0, 1, 2, 3 and Prob: \(\frac{1}{45}\), \(\frac{8}{45}\), \(\frac{20}{45}\), \(\frac{16}{45}\) | B1 | Probability distribution table with correct outcomes with at least one probability, allow extra outcome values if probability of zero stated |
| B1 | 2 of P(0), P(1) and P(3) correct | |
| B1 FT | 3 or 4 probabilities sum to 1 with P(2) correct | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Var}(X) = \frac{0^2 \times 1 + 1^2 \times 8 + 2^2 \times 20 + 3^2 \times 16}{45} - \left(\frac{32}{15}\right)^2\) | M1 | Substitute their attempts at scores in correct variance formula, must have \(- \text{mean}^2\) (FT if calculated); must be at least 2 non-zero values |
| \(= \frac{8}{45} + \frac{80}{45} + \frac{144}{45} - \left(\frac{32}{15}\right)^2\) | ||
| \(\frac{136}{225}\) or \(0.604\) | A1 | |
| 2 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| HHT: $\frac{2}{3} \times \frac{2}{3} \times \frac{1}{5} = \frac{4}{45}$ | M1 | One 3 factor probability with 3, 3, 5 as denominators |
| HTH: $\frac{2}{3} \times \frac{1}{3} \times \frac{4}{5} = \frac{8}{45}$ | M1 | 3 factor probabilities for 2 or 3 correct scenarios added, no incorrect scenarios |
| THH: $\frac{1}{3} \times \frac{2}{3} \times \frac{4}{5} = \frac{8}{45}$ | | |
| Total $= \frac{20}{45} = \frac{4}{9}$ | A1 | AG, Total of 3 products with clear context |
| | **3** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with $x$: 0, 1, 2, 3 and Prob: $\frac{1}{45}$, $\frac{8}{45}$, $\frac{20}{45}$, $\frac{16}{45}$ | B1 | Probability distribution table with correct outcomes with at least one probability, allow extra outcome values if probability of zero stated |
| | B1 | 2 of P(0), P(1) and P(3) correct |
| | B1 FT | 3 or 4 probabilities sum to 1 with P(2) correct |
| | **3** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Var}(X) = \frac{0^2 \times 1 + 1^2 \times 8 + 2^2 \times 20 + 3^2 \times 16}{45} - \left(\frac{32}{15}\right)^2$ | M1 | Substitute their attempts at scores in correct variance formula, must have $- \text{mean}^2$ (FT if calculated); must be at least 2 non-zero values |
| $= \frac{8}{45} + \frac{80}{45} + \frac{144}{45} - \left(\frac{32}{15}\right)^2$ | | |
| $\frac{136}{225}$ or $0.604$ | A1 | |
| | **2** | |6 Three coins $A , B$ and $C$ are each thrown once.
\begin{itemize}
\item Coins $A$ and $B$ are each biased so that the probability of obtaining a head is $\frac { 2 } { 3 }$.
\item Coin $C$ is biased so that the probability of obtaining a head is $\frac { 4 } { 5 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability of obtaining exactly 2 heads and 1 tail is $\frac { 4 } { 9 }$.\\
\end{itemize}
The random variable $X$ is the number of heads obtained when the three coins are thrown.
\item Draw up the probability distribution table for $X$.
\item Given that $\mathrm { E } ( X ) = \frac { 32 } { 15 }$, find $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q6 [8]}}