| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Multiple independent trials or dice |
| Difficulty | Easy -1.2 This is a straightforward S1 question testing basic discrete uniform distribution properties (naming, mean/variance formulas) and simple probability calculations with independent events. Part (d) requires systematic enumeration of outcomes summing to 16, but this is routine combinatorial listing rather than complex problem-solving. All parts use standard textbook methods with no novel insight required. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Discrete uniform | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=x) = \frac{1}{6}, \quad x = 1, 2, \ldots, 6\) | B1 | |
| \(\therefore E(X) = \Sigma x \, P(X=x) = \frac{1}{6} + \frac{2}{6} + \ldots + \frac{6}{6} = \frac{21}{6} = 3.5\) | B1 M1 | |
| \(\text{Var}(X) = \Sigma x^2 P(X=x) - \{E(X)\}^2\) | ||
| \(= \frac{1}{6} + \frac{4}{6} + \ldots + \frac{36}{6} - \left(\frac{21}{6}\right)^2 = 2.91666\ldots\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{three 6s}) = \left(\frac{1}{6}\right)^3 = \frac{1}{216}\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(16 \Rightarrow (6,5,5); (5,6,5); (5,5,6)\) | B1 B1 | |
| \((6,6,4); (6,4,6); (4,6,6)\) | B1 B1 | |
| \((6,6,6); (6,4,6); (4,6,6)\) | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(16) = \frac{6}{216} = \frac{1}{36}\) | M1 A1 | (2 marks) |
| (12 marks) |
**(a)**
| Discrete uniform | B1 | (1 mark) |
**(b)**
| $P(X=x) = \frac{1}{6}, \quad x = 1, 2, \ldots, 6$ | B1 | |
| $\therefore E(X) = \Sigma x \, P(X=x) = \frac{1}{6} + \frac{2}{6} + \ldots + \frac{6}{6} = \frac{21}{6} = 3.5$ | B1 M1 | |
| $\text{Var}(X) = \Sigma x^2 P(X=x) - \{E(X)\}^2$ | | |
| $= \frac{1}{6} + \frac{4}{6} + \ldots + \frac{36}{6} - \left(\frac{21}{6}\right)^2 = 2.91666\ldots$ | A1 | (3 marks) |
**(c)**
| $P(\text{three 6s}) = \left(\frac{1}{6}\right)^3 = \frac{1}{216}$ | M1 A1 | (2 marks) |
**(d)**
| $16 \Rightarrow (6,5,5); (5,6,5); (5,5,6)$ | B1 B1 | |
| | $(6,6,4); (6,4,6); (4,6,6)$ | B1 B1 | |
| | $(6,6,6); (6,4,6); (4,6,6)$ | | (4 marks) |
**(e)**
| $P(16) = \frac{6}{216} = \frac{1}{36}$ | M1 A1 | (2 marks) |
| | | (12 marks) |
---5. The random variable $X$ represents the number on the uppermost face when a fair die is thrown.
\begin{enumerate}[label=(\alph*)]
\item Write down the name of the probability distribution of $X$.
\item Calculate the mean and the variance of $X$.
Three fair dice are thrown and the numbers on the uppermost faces are recorded.
\item Find the probability that all three numbers are 6 .
\item Write down all the different ways of scoring a total of 16 when the three numbers are added together.
\item Find the probability of scoring a total of 16 .
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2003 Q5 [12]}}