| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from given formula |
| Difficulty | Moderate -0.3 This is a straightforward discrete probability distribution question requiring systematic enumeration of outcomes (part a), using the sum of probabilities equals 1 (part b), and applying standard variance formula (part c). While it involves multiple steps and careful bookkeeping across three games, the techniques are all standard S1 material with no novel problem-solving required. Slightly easier than average due to being methodical rather than conceptually challenging. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04a Discrete probability distributions |
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm { P } ( \mathrm { X } = \mathrm { x } )\) | 0.114 | 0.207 | 0.285 | 0.125 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=2) = P(\text{WLL or DDL})\) \(= 0.5 \times 0.2^2 \times {}^3C_1 + 0.3^2 \times 0.2 \times {}^3C_1 = [0.06 + 0.054]\) | M1 | \(0.5 \times 0.2^2 \times {}^3C_a\ (or\ 3) + x;\ 0 < x < 1;\ a = 1, 2.\) Or \(0.3^2 \times 0.2 \times {}^3C_b\ (or\ 3) + y;\ 0 < y < 1;\ b = 1, 2.\) Or \(0.5 \times 0.2^2 \times a + 0.3^2 \times 0.2 \times b;\ a,\ b = 1, 2, 3.\) |
| \(= 0.114\) | A1 | AG. Fully correct solutions with outcomes identified and linked to appropriate probabilities. Condone \(2 = W,\ 1 = D,\ 0 = L\). Probabilities alone do not identify outcomes. If individual scenarios are identified, separate calculations must correspond to the order. |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): 0, 1, 2, 3, 4, 5, 6 | B1 | One additional correct probability in table or clearly identified. |
| \(P(X=x)\): 0.008 \(\left(\frac{1}{125}\right)\), 0.0369 \(\left(\frac{9}{250}\right)\), 0.114, 0.207, 0.285, 0.225 \(\left(\frac{9}{40}\right)\), 0.125 | B1 | A second additional correct probability in table or clearly identified. |
| B1 | Final correct probability, all probabilities in table. If 0/3 scored, SC B1 for three additional probabilities in table that sum to 0.269 exactly. | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = [0.008\times0]+0.036\times1+0.114\times2+0.207\times3+0.285\times4+0.225\times5+0.125\times6\) \(= [0]+0.036+0.228+0.621+1.140+1.125+0.750\ [=3.9]\) Or \(E(X) = 3(0.5\times2 + 0.3\times1)\ [=3.9]\) | M1 | Accept unsimplified expression. May be calculated in the variance, FT *their* table with probabilities \(0 < p < 1\), that sum to 1. FT acceptable at the bold partially evaluated stage. |
| \(\text{Var}(X) = [0.008\times0^2+]0.036\times1^2+0.114\times2^2+0.207\times3^2+0.285\times4^2+0.225\times5^2+0.125\times6^2 - \textit{their}\ 3.9^2\) | M1 | Appropriate variance formula using *their* \((E(X))^2\) value. FT *their* table with probabilities \(0 < p < 1\), that may not sum to 1. |
| \(= [17.04 - 3.9^2] = 1.83\) | A1 | Cao. Condone \(\frac{183}{100}\) |
| Total | 3 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=2) = P(\text{WLL or DDL})$ $= 0.5 \times 0.2^2 \times {}^3C_1 + 0.3^2 \times 0.2 \times {}^3C_1 = [0.06 + 0.054]$ | M1 | $0.5 \times 0.2^2 \times {}^3C_a\ (or\ 3) + x;\ 0 < x < 1;\ a = 1, 2.$ Or $0.3^2 \times 0.2 \times {}^3C_b\ (or\ 3) + y;\ 0 < y < 1;\ b = 1, 2.$ Or $0.5 \times 0.2^2 \times a + 0.3^2 \times 0.2 \times b;\ a,\ b = 1, 2, 3.$ |
| $= 0.114$ | A1 | AG. Fully correct solutions with outcomes identified and linked to appropriate probabilities. Condone $2 = W,\ 1 = D,\ 0 = L$. Probabilities alone do not identify outcomes. If individual scenarios are identified, separate calculations must correspond to the order. |
| **Total** | **2** | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: 0, 1, 2, 3, 4, 5, 6 | B1 | One additional correct probability in table or clearly identified. |
| $P(X=x)$: 0.008 $\left(\frac{1}{125}\right)$, 0.0369 $\left(\frac{9}{250}\right)$, 0.114, 0.207, 0.285, 0.225 $\left(\frac{9}{40}\right)$, 0.125 | B1 | A second additional correct probability in table or clearly identified. |
| | B1 | Final correct probability, all probabilities in table. If 0/3 scored, **SC B1** for three additional probabilities in table that sum to 0.269 exactly. |
| **Total** | **3** | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = [0.008\times0]+0.036\times1+0.114\times2+0.207\times3+0.285\times4+0.225\times5+0.125\times6$ $= [0]+0.036+0.228+0.621+1.140+1.125+0.750\ [=3.9]$ Or $E(X) = 3(0.5\times2 + 0.3\times1)\ [=3.9]$ | M1 | Accept unsimplified expression. May be calculated in the variance, FT *their* table with probabilities $0 < p < 1$, that sum to 1. FT acceptable at the bold partially evaluated stage. |
| $\text{Var}(X) = [0.008\times0^2+]0.036\times1^2+0.114\times2^2+0.207\times3^2+0.285\times4^2+0.225\times5^2+0.125\times6^2 - \textit{their}\ 3.9^2$ | M1 | Appropriate variance formula using *their* $(E(X))^2$ value. FT *their* table with probabilities $0 < p < 1$, that may not sum to 1. |
| $= [17.04 - 3.9^2] = 1.83$ | A1 | Cao. Condone $\frac{183}{100}$ |
| **Total** | **3** | |
---5 Anil is taking part in a tournament. In each game in this tournament, players are awarded 2 points for a win, 1 point for a draw and 0 points for a loss. For each of Anil's games, the probabilities that he will win, draw or lose are $0.5,0.3$ and 0.2 respectively. The results of the games are all independent of each other.
The random variable $X$ is the total number of points that Anil scores in his first 3 games in the tournament.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 2 ) = 0.114$.
\item Complete the probability distribution table for $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { x } )$ & & & 0.114 & 0.207 & 0.285 & & 0.125 \\
\hline
\end{tabular}
\end{center}
\item Find the value of $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q5 [8]}}