| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from constructed distribution |
| Difficulty | Moderate -0.8 This is a straightforward expectation and variance calculation from a given probability distribution table. Part (i) requires simple logical reasoning (if 3 are correct, the 4th must be too), part (ii) is basic counting (1 arrangement out of 4! = 24), and part (iii) is direct application of standard formulas E(X) = Σxp(x) and Var(X) = E(X²) - [E(X)]². The distribution is provided, so no construction is needed—just arithmetic with fractions. This is easier than average A-level statistics work. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 3 } { 8 }\) | \(\frac { 1 } { 3 }\) | \(\frac { 1 } { 4 }\) | 0 | \(\frac { 1 } { 24 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| If 3 letters are in the correct envelope, the 4th must also be correct | B1 | Accept equivalent explanation |
| Answer | Marks |
|---|---|
| \(P(X=4) = \frac{1}{4!} = \frac{1}{24}\) since there is only one arrangement where all are correct out of \(4! = 24\) equally likely arrangements | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 0 \times \frac{3}{8} + 1 \times \frac{1}{3} + 2 \times \frac{1}{4} + 3 \times 0 + 4 \times \frac{1}{24}\) | M1 | |
| \(= 0 + \frac{1}{3} + \frac{1}{2} + 0 + \frac{1}{6} = 1\) | A1 | cao |
| \(E(X^2) = 0 + \frac{1}{3} + 1 + 0 + \frac{16}{24} = \frac{1}{3} + 1 + \frac{2}{3} = 2\) | M1 | |
| \(Var(X) = E(X^2) - [E(X)]^2 = 2 - 1 = 1\) | M1 A1 |
# Question 2:
**(i)**
If 3 letters are in the correct envelope, the 4th must also be correct | B1 | Accept equivalent explanation
**(ii)**
$P(X=4) = \frac{1}{4!} = \frac{1}{24}$ since there is only one arrangement where all are correct out of $4! = 24$ equally likely arrangements | M1 A1 |
**(iii)**
$E(X) = 0 \times \frac{3}{8} + 1 \times \frac{1}{3} + 2 \times \frac{1}{4} + 3 \times 0 + 4 \times \frac{1}{24}$ | M1 |
$= 0 + \frac{1}{3} + \frac{1}{2} + 0 + \frac{1}{6} = 1$ | A1 | cao
$E(X^2) = 0 + \frac{1}{3} + 1 + 0 + \frac{16}{24} = \frac{1}{3} + 1 + \frac{2}{3} = 2$ | M1 |
$Var(X) = E(X^2) - [E(X)]^2 = 2 - 1 = 1$ | M1 A1 |
---2 Four letters are taken out of their envelopes for signing. Unfortunately they are replaced randomly, one in each envelope.
The probability distribution for the number of letters, $X$, which are now in the correct envelope is given in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 3 } { 8 }$ & $\frac { 1 } { 3 }$ & $\frac { 1 } { 4 }$ & 0 & $\frac { 1 } { 24 }$ \\
\hline
\end{tabular}
\end{center}
(i) Explain why the case $X = 3$ is impossible.\\
(ii) Explain why $\mathrm { P } ( X = 4 ) = \frac { 1 } { 24 }$.\\
(iii) Calculate $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 2006 Q2 [8]}}