| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from combinatorial selection |
| Difficulty | Moderate -0.8 This is a straightforward combinatorial probability question requiring basic counting (choosing 3 from 6) and standard expectation/variance calculations from a given distribution table. Part (i) is pure recall of combinations formula, and part (ii) involves routine application of E(X) and Var(X) formulas with simple arithmetic. No problem-solving insight needed, just mechanical application of standard S1 techniques. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p) |
| \(r\) | 15 | 1010 | 2005 | 3000 |
| \(\mathrm { P } ( X = r )\) | 0.05 | 0.45 | 0.45 | 0.05 |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1 | A1 | NB |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1 | A1 | M1 |
## Part (i)
**Answer:**
$$P(X = 15) = \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} = \frac{6}{120} = \frac{1}{20} = 0.05$$
Or:
$$\frac{1}{{}^6C_3} = \frac{1}{20} = 0.05$$
Or:
$$\frac{3! \times 3!}{6!} = \frac{1}{20} = 0.05$$
**Marks:** M1 | A1 | NB | [2]
**Guidance:**
- M1: For product of three correct fractions
- A1: NB ANSWER GIVEN. Full marks for $31 \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} = \frac{6}{120} = 0.05$
- NB: $1 - (0.45 + 0.45 + 0.05) = 0.05$ scores M0A0. Allow 3 × 2 in place of 3!. SC1 for $6 \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} = \frac{120}{120} = 0.05$
## Part (ii)
**Answer:**
$$E(X) = (15 \times 0.05) + (1010 \times 0.45) + (2005 \times 0.45) + (3000 \times 0.05)$$
$$= 1507.5 \text{ so } 1508 \text{ (4sf)}$$
$$E(X^2) = (15^2 \times 0.05) + (1010^2 \times 0.45) + (2005^2 \times 0.45) + (3000^2 \times 0.05)$$
$$= 2718067.5$$
$$\text{Var}(X) = 2718067.5 - (1507.5)^2 = 445511.25 \approx 445500 \text{ (4sf)}$$
**Marks:** M1 | A1 | M1 | M1 | A1 | [5]
**Guidance:**
- M1: For $\sum rp$ (at least 3 terms correct)
- A1: CAO. Allow 1507, 1510, 1507.5, 1507.50 or 3015/2
- M1: For $\sum r^2 p$ (at least 3 terms correct)
- M1: dep for – their E(X)². Use of E(X-μ)² gets M1 for attempt at (x-μ)²; should see (-1492.5)², (-497.5)², 497.5², 1492.5², (if E(X) wrong FT their E(X))) (all 4 correct for M1), then M1 for $\sum p(x-\mu)^2$ (at least 3 terms correct with their probabilities). Division by 4 or other spurious value at end gives max M1M1M1A0, or M1A0M1M1A0 if E(X) also divided by 4.
- A1: FT their E(X) provided Var(X) > 0 (and of course E(X²) is correct). Allow 446000
---4 Martin has won a competition. For his prize he is given six sealed envelopes, of which he is allowed to open exactly three and keep their contents. Three of the envelopes each contain $\pounds 5$ and the other three each contain $\pounds 1000$. Since the envelopes are identical on the outside, he chooses three of them at random. Let $\pounds X$ be the total amount of money that he receives in prize money.\\
(i) Show that $\mathrm { P } ( X = 15 ) = 0.05$.
The probability distribution of $X$ is given in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$r$ & 15 & 1010 & 2005 & 3000 \\
\hline
$\mathrm { P } ( X = r )$ & 0.05 & 0.45 & 0.45 & 0.05 \\
\hline
\end{tabular}
\end{center}
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 2013 Q4 [7]}}