| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Find cumulative distribution F(x) |
| Difficulty | Moderate -0.3 This is a straightforward S1 question testing standard discrete probability distribution calculations. Parts (a)-(d) are routine textbook exercises requiring direct application of formulas for expectation, cumulative distribution, and variance. Part (e) requires listing outcomes for two games summing to ≥4, which is mechanical enumeration. Slightly easier than average due to small sample space and no conceptual challenges. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04b Linear combinations: of normal distributions |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | 0.4 | 0.3 | 0.2 | 0.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 0 \times 0.4 + 1 \times 0.3 + \ldots + 3 \times 0.1 = 1\) | M1, A1 (2) | M1 for at least 3 terms seen. Correct answer only scores M1A1. Dividing by \(k \neq 1\) is M0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(1.5) = [P(X \leq 1.5) =] P(X \leq 1) = 0.4 + 0.3 = 0.7\) | M1, A1 (2) | M1 for \(F(1.5) = P(X \leq 1)\). Beware: \(2 \times 0.2 + 3 \times 0.1 = 0.7\) scores M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.3 + \ldots + 3^2 \times 0.1 = 2\) | M1, A1 | 1st M1 for at least 2 non-zero terms; \(E(X^2) = 2\) alone is M0; 1st A1 for answer of 2 or fully correct expression |
| \(\text{Var}(X) = 2 - 1^2 = 1\) | M1, A1cso (4) | 2nd M1 for \(-\mu^2\); 2nd A1 for fully correct solution with no incorrect working, both Ms required |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(5 - 3X) = (-3)^2 \text{Var}(X) = 9\) | M1, A1 (2) | M1 for use of correct formula; \(-3^2 \text{Var}(X)\) is M0 unless final answer \(> 0\). Can follow through their \(\text{Var}(X)\) for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | Cases | Probability |
| 4 | \((X=3)\cap(X=1)\) | \(0.1 \times 0.3 = 0.03\) |
| 4 | \((X=1)\cap(X=3)\) | \(0.3 \times 0.1 = 0.03\) |
| 4 | \((X=2)\cap(X=2)\) | \(0.2 \times 0.2 = 0.04\) |
| 5 | \((X=3)\cap(X=2)\) | \(0.1 \times 0.2 = 0.02\) |
| 5 | \((X=2)\cap(X=3)\) | \(0.2 \times 0.1 = 0.02\) |
| 6 | \((X=3)\cap(X=3)\) | \(0.1 \times 0.1 = 0.01\) |
| B1B1B1 | 1st B1 for all cases listed for total of 4 or 5 or 6; (2,2) counted twice for total of 4 is B0. 2nd B1 for all cases listed for 2 totals. 3rd B1 for complete list of all 6 cases | |
| Total probability \(= 0.03 + 0.03 + 0.04 + 0.02 + 0.02 + 0.01 = 0.15\) | M1, A1 (6) | M1 for one correct pair of probabilities multiplied; A1 for 0.15 or exact equivalent only |
## Question 3:
**(a)**
$E(X) = 0 \times 0.4 + 1 \times 0.3 + \ldots + 3 \times 0.1 = 1$ | M1, A1 (2) | M1 for at least 3 terms seen. Correct answer only scores M1A1. Dividing by $k \neq 1$ is M0
**(b)**
$F(1.5) = [P(X \leq 1.5) =] P(X \leq 1) = 0.4 + 0.3 = 0.7$ | M1, A1 (2) | M1 for $F(1.5) = P(X \leq 1)$. Beware: $2 \times 0.2 + 3 \times 0.1 = 0.7$ scores M0A0
**(c)**
$E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.3 + \ldots + 3^2 \times 0.1 = 2$ | M1, A1 | 1st M1 for at least 2 non-zero terms; $E(X^2) = 2$ alone is M0; 1st A1 for answer of 2 or fully correct expression
$\text{Var}(X) = 2 - 1^2 = 1$ | M1, A1cso (4) | 2nd M1 for $-\mu^2$; 2nd A1 for fully correct solution with no incorrect working, both Ms required
**(d)**
$\text{Var}(5 - 3X) = (-3)^2 \text{Var}(X) = 9$ | M1, A1 (2) | M1 for use of correct formula; $-3^2 \text{Var}(X)$ is M0 unless final answer $> 0$. Can follow through their $\text{Var}(X)$ for M1
**(e)**
| Total | Cases | Probability |
|-------|-------|-------------|
| 4 | $(X=3)\cap(X=1)$ | $0.1 \times 0.3 = 0.03$ |
| 4 | $(X=1)\cap(X=3)$ | $0.3 \times 0.1 = 0.03$ |
| 4 | $(X=2)\cap(X=2)$ | $0.2 \times 0.2 = 0.04$ |
| 5 | $(X=3)\cap(X=2)$ | $0.1 \times 0.2 = 0.02$ |
| 5 | $(X=2)\cap(X=3)$ | $0.2 \times 0.1 = 0.02$ |
| 6 | $(X=3)\cap(X=3)$ | $0.1 \times 0.1 = 0.01$ |
| B1B1B1 | 1st B1 for all cases listed for total of 4 or 5 or 6; (2,2) counted twice for total of 4 is B0. 2nd B1 for all cases listed for 2 totals. 3rd B1 for complete list of all 6 cases
Total probability $= 0.03 + 0.03 + 0.04 + 0.02 + 0.02 + 0.01 = 0.15$ | M1, A1 (6) | M1 for one correct pair of probabilities multiplied; A1 for 0.15 or exact equivalent only
---3. When Rohit plays a game, the number of points he receives is given by the discrete random variable $X$ with the following probability distribution.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & 0.4 & 0.3 & 0.2 & 0.1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( X )$.
\item Find $\mathrm { F } ( 1.5 )$.
\item Show that $\operatorname { Var } ( X ) = 1$
\item Find $\operatorname { Var } ( 5 - 3 X )$.
Rohit can win a prize if the total number of points he has scored after 5 games is at least 10. After 3 games he has a total of 6 points. You may assume that games are independent.
\item Find the probability that Rohit wins the prize.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2009 Q3 [16]}}