| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probability distribution from formula |
| Difficulty | Moderate -0.3 This is a straightforward S1 question requiring basic probability distribution properties. Part (i) involves simple substitution and using ΣP=1 to find k. Part (ii) is standard expectation calculation. Part (iii) requires systematic enumeration of outcomes totaling 16, which is routine but slightly tedious. All techniques are standard textbook exercises with no novel insight required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.04a Discrete probability distributions |
| \(r\) | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm { P } ( X = r )\) | \(k\) | \(11 k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=r)\) values: \(k, 3k, 5k, 7k, 9k, 11k\) | B1 | B1 for \(3k, 5k, 7k, 9k\) |
| \(36k = 1\), so \(k = \frac{1}{36}\) | M1, A1 CAO | M1 for sum of six multiples of \(k=1\); answer MUST BE FRACTION IN SIMPLEST FORM |
| Answer | Marks | Guidance |
|---|---|---|
| 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, A1 CAO | M1 for \(\sum rp\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=16) = 6\times\left(\frac{1}{6}\right)^3 = \frac{6}{216} = \frac{1}{36}\) | M1, M1, A1 CAO | M1 for \(6\times\); M1 indep for \(\left(\frac{1}{6}\right)^3\) |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=r)$ values: $k, 3k, 5k, 7k, 9k, 11k$ | B1 | B1 for $3k, 5k, 7k, 9k$ |
| $36k = 1$, so $k = \frac{1}{36}$ | M1, A1 CAO | M1 for sum of six multiples of $k=1$; **answer MUST BE FRACTION IN SIMPLEST FORM** |
## Part (ii)
| 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, A1 CAO | M1 for $\sum rp$ |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=16) = 6\times\left(\frac{1}{6}\right)^3 = \frac{6}{216} = \frac{1}{36}$ | M1, M1, A1 CAO | M1 for $6\times$; M1 indep for $\left(\frac{1}{6}\right)^3$ |
---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 2007 Q4 [8]}}