| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.8 This is a straightforward S1 question requiring basic probability distribution properties. Part (i) involves simple substitution and solving ΣP(X=r)=1 for k. Part (ii) is routine expectation calculation. Part (iii) requires counting outcomes for a sum of 16 from three dice, which is slightly more involved but still standard. Overall, this is easier than average A-level, requiring only mechanical application of formulas with no conceptual challenges. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(r\) | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm { P } ( X = r )\) | \(k\) | \(11 k\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r\) values: 1, 2, 3, 4, 5, 6 with \(P(X=r)\): \(k, 3k, 5k, 7k, 9k, 11k\) | B1 | for \(3k, 5k, 7k, 9k\) |
| \(36k = 1\), so \(k = \frac{1}{36}\) | M1 | for sum of six multiples of \(k = 1\) |
| A1 CAO | MUST BE FRACTION IN SIMPLEST FORM |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 1\times\frac{1}{36} + 2\times\frac{3}{36} + 3\times\frac{5}{36} + 4\times\frac{7}{36} + 5\times\frac{9}{36} + 6\times\frac{11}{36} = \frac{161}{36} = 4.47\) | M1 | for \(\Sigma rp\) |
| A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=16) = 6\times\left(\frac{1}{6}\right)^3 = \frac{6}{216} = \frac{1}{36}\) | M1 | for \(6\times\) |
| M1 indep | for \(\left(\frac{1}{6}\right)^3\) | |
| A1 CAO |
## Question 4:
### Part (i)
| $r$ values: 1, 2, 3, 4, 5, 6 with $P(X=r)$: $k, 3k, 5k, 7k, 9k, 11k$ | B1 | for $3k, 5k, 7k, 9k$ |
| $36k = 1$, so $k = \frac{1}{36}$ | M1 | for sum of six multiples of $k = 1$ |
| | A1 CAO | **MUST BE FRACTION IN SIMPLEST FORM** |
### Part (ii)
| $E(X) = 1\times\frac{1}{36} + 2\times\frac{3}{36} + 3\times\frac{5}{36} + 4\times\frac{7}{36} + 5\times\frac{9}{36} + 6\times\frac{11}{36} = \frac{161}{36} = 4.47$ | M1 | for $\Sigma rp$ |
| | A1 CAO | |
### Part (iii)
| $P(X=16) = 6\times\left(\frac{1}{6}\right)^3 = \frac{6}{216} = \frac{1}{36}$ | M1 | for $6\times$ |
| | M1 indep | for $\left(\frac{1}{6}\right)^3$ |
| | A1 CAO | |
---4 A fair six-sided die is rolled twice. The random variable $X$ represents the higher of the two scores. The probability distribution of $X$ is given by the formula
$$\mathrm { P } ( X = r ) = k ( 2 r - 1 ) \text { for } r = 1,2,3,4,5,6$$
(i) Copy and complete the following probability table and hence find the exact value of $k$, giving your answer as a fraction in its simplest form.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( X = r )$ & $k$ & & & & & $11 k$ \\
\hline
\end{tabular}
\end{center}
(ii) Find the mean of $X$.
A fair six-sided die is rolled three times.\\
(iii) Find the probability that the total score is 16 .
\hfill \mbox{\textit{OCR MEI S1 Q4 [8]}}