| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.3 This is a standard two-equation, two-unknown problem requiring routine application of probability axioms (sum = 1) and expectation formula. The algebra is straightforward with no conceptual challenges, making it slightly easier than average for A-level. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a+b,\ a+2b,\ a+3b\) | B1 | All three seen |
| \(a+b+a+2b+a+3b=1\) oe | B1dep | Must see this line oe before final answer or "Probabilities add up to 1" oe stated; Must include "\(=1\)" |
| \((3a+6b=1\ \mathbf{AG})\) | ||
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a+b+2(a+2b)+3(a+3b) = \frac{5}{3}\) | M1 | ft their probs |
| \(6a+14b=\frac{5}{3}\) or \(18a+42b=5\) | A1f | or any correct three term equation, ft their probs |
| e.g. \(6 \times \frac{1-6b}{3} + 14b = \frac{5}{3}\) or \(2b = -\frac{1}{3}\) | ||
| or \(6a + 14 \times \frac{1-3a}{6} = \frac{5}{3}\) or \(3a=2\) | A1 | or any correct equation in \(a\) or \(b\) only, cao |
| \(a = \frac{2}{3},\ b = -\frac{1}{6}\) | A1 | cao |
| [4] |
# Question 9:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a+b,\ a+2b,\ a+3b$ | B1 | All three seen |
| $a+b+a+2b+a+3b=1$ oe | B1dep | Must see this line oe before final answer or "Probabilities add up to 1" oe stated; Must include "$=1$" |
| $(3a+6b=1\ \mathbf{AG})$ | | |
| **[2]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a+b+2(a+2b)+3(a+3b) = \frac{5}{3}$ | M1 | ft their probs |
| $6a+14b=\frac{5}{3}$ or $18a+42b=5$ | A1f | or any correct three term equation, ft their probs |
| e.g. $6 \times \frac{1-6b}{3} + 14b = \frac{5}{3}$ or $2b = -\frac{1}{3}$ | | |
| or $6a + 14 \times \frac{1-3a}{6} = \frac{5}{3}$ or $3a=2$ | A1 | or any correct equation in $a$ or $b$ only, cao |
| $a = \frac{2}{3},\ b = -\frac{1}{6}$ | A1 | cao |
| **[4]** | | |9 The random variable $X$ has probability distribution given by
$$\mathrm { P } ( X = x ) = a + b x \quad \text { for } x = 1,2 \text { and } 3 ,$$
where $a$ and $b$ are constants.\\
(i) Show that $3 a + 6 b = 1$.\\
(ii) Given that $\mathrm { E } ( X ) = \frac { 5 } { 3 }$, find $a$ and $b$.
\hfill \mbox{\textit{OCR S1 2015 Q9 [6]}}