| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Probability distributions with parameters |
| Difficulty | Standard +0.8 This S4 joint probability distribution question requires systematic enumeration of all outcomes, finding a normalizing constant, computing marginal/conditional probabilities, and testing independence. While methodical rather than conceptually deep, the multi-part nature, conditional expectation calculation, and independence test make it moderately challenging—above average but accessible to well-prepared students. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Table with \(N\): 0,1,2 and \(R\): 0,1,2 giving probabilities \(0, c, 2c, 2c, 3c, 4c, 4c, 5c, 6c\) | M1 | Calculate 9 probs in terms of \(c\) |
| Total \(27c = 1\) | ||
| \(c = \frac{1}{27}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9c/27c\) | M1 | Marginal probability |
| \(= \frac{1}{3}\) | A1ft [2] | AEF; ft \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(N + R > 2) = 15c/27c = \frac{5}{9}\) | M1, A1ft [2] | AEF; ft \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(R=2) = \frac{15}{27}\) | M1 | Using conditional probabilities |
| \(P(N \mid R=2)\): \(p_0 = \frac{4}{15}\), \(p_1 = \frac{1}{3}\), \(p_2 = \frac{2}{5}\) | A1ft | One value; ft values in (i) |
| \(E(N \mid R=2) = 1 \times \frac{1}{3} + 2 \times \frac{2}{5}\) | A1ft | All values |
| \(= \frac{17}{15}\) | A1 [4] | Or 1.13 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Eg \(P(N=0 \text{ and } R=0) = 0\) | M1 | Or from conditional probs; M0 from \(N=1\) with \(R=1\) or 2 |
| \(P(N=0) \times P(R=0) = \frac{6}{27} \times \frac{3}{27} \neq 0\) | All correct | |
| So \(N\) and \(R\) are not independent | A1 [2] |
## Question 6:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Table with $N$: 0,1,2 and $R$: 0,1,2 giving probabilities $0, c, 2c, 2c, 3c, 4c, 4c, 5c, 6c$ | M1 | Calculate 9 probs in terms of $c$ |
| Total $27c = 1$ | | |
| $c = \frac{1}{27}$ | A1 **[3]** | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9c/27c$ | M1 | Marginal probability |
| $= \frac{1}{3}$ | A1ft **[2]** | AEF; ft $c$ |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(N + R > 2) = 15c/27c = \frac{5}{9}$ | M1, A1ft **[2]** | AEF; ft $c$ |
### Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(R=2) = \frac{15}{27}$ | M1 | Using conditional probabilities |
| $P(N \mid R=2)$: $p_0 = \frac{4}{15}$, $p_1 = \frac{1}{3}$, $p_2 = \frac{2}{5}$ | A1ft | One value; ft values in (i) |
| $E(N \mid R=2) = 1 \times \frac{1}{3} + 2 \times \frac{2}{5}$ | A1ft | All values |
| $= \frac{17}{15}$ | A1 **[4]** | Or 1.13 |
### Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Eg $P(N=0 \text{ and } R=0) = 0$ | M1 | Or from conditional probs; M0 from $N=1$ with $R=1$ or 2 |
| $P(N=0) \times P(R=0) = \frac{6}{27} \times \frac{3}{27} \neq 0$ | | All correct |
| So $N$ and $R$ are not independent | A1 **[2]** | |
---6 Nuts and raisins occur in randomly chosen squares of a particular brand of chocolate. The numbers of nuts and raisins are denoted by $N$ and $R$ respectively and the joint probability distribution of $N$ and $R$ is given by
$$f ( n , r ) = \begin{cases} c ( n + 2 r ) & n = 0,1,2 \text { and } r = 0,1,2 \\ 0 & \text { otherwise } \end{cases}$$
where $c$ is a constant.\\
(i) Find the value of $c$.\\
(ii) Find the probability that there is exactly one nut in a randomly chosen square.\\
(iii) Find the probability that the total number of nuts and raisins in a randomly chosen square is more than 2 .\\
(iv) For squares in which there are 2 raisins, find the mean number of nuts.\\
(v) Determine whether $N$ and $R$ are independent.
\hfill \mbox{\textit{OCR S4 2010 Q6 [13]}}