| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Joint distribution with marginal/conditional probabilities |
| Difficulty | Standard +0.3 This is a straightforward joint probability distribution question requiring verification of given probabilities using combinatorics, calculation of expectations from marginal distributions, and basic understanding of independence. All parts follow standard procedures with no novel insight required, making it slightly easier than average for S4 level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(L\) | |||
| 0 | 1 | 2 | |
| 0 | \(\frac { 1 } { 42 }\) | \(\frac { 16 } { 105 }\) | \(\frac { 4 } { 21 }\) |
| 1 | \(\frac { 2 } { 15 }\) | \(\frac { 16 } { 45 }\) | 0 |
| 2 | \(\frac { 13 } { 90 }\) | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(L=0\), \(C=1\): choose 1C from 14 and 1 from 6 Others | M1 | Or \(\frac{14}{36} \times \frac{6}{35} \times 2\) |
| \(14 \times 6\ /\ ^{36}C_2 = \frac{2}{15}\) AG | A1 | |
| \(L=1\), \(C=1\): choose 1 from 16, 1 from 14 | M1 | Or \(\frac{14}{36} \times \frac{16}{35} \times 2\) |
| \(16 \times 14\ /\ ^{36}C_2 = \frac{16}{45}\) AG | A1 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Marginal C probs: \(\frac{11}{30}\), \(\frac{22}{45}\), \(\frac{13}{90}\) | B1 | AEF |
| \(E(C) = \frac{22}{45} + \frac{26}{90} = \frac{35}{45} = \frac{7}{9}\) | M1 | |
| A1 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: \(2 \times \frac{1}{42} + \frac{2}{15} + \frac{16}{105}\) | M1 | Other: 0, 1, 2 M1 |
| \(E(L) = \frac{8}{9}\), \(E(O) = 2 - \frac{15}{9}\) | A1 | \(p\): \(\frac{29}{42}\), \(\frac{2}{7}\), \(\frac{1}{42}\) A1 |
| \(= \frac{1}{3}\) | A1 3 | \(E(O) = \frac{2}{7} + \frac{2}{42} = \frac{1}{3}\) A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: Argument | B2 | e.g. The more \(L\)s the fewer \(C\)s |
| OR: Use idea that for independence \(P(L \cap C) = P(L)P(C)\) | M1A1 | OR Use conditional probability |
| Conclude that covariance is non-zero | B1 3 | OR \(\text{Cov}(L,C) = -\frac{136}{405}\) M1A1; \(L\),\(C\) not indep B1 |
| [13] |
## Question 6:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $L=0$, $C=1$: choose 1C from 14 and 1 from 6 Others | M1 | Or $\frac{14}{36} \times \frac{6}{35} \times 2$ |
| $14 \times 6\ /\ ^{36}C_2 = \frac{2}{15}$ AG | A1 | |
| $L=1$, $C=1$: choose 1 from 16, 1 from 14 | M1 | Or $\frac{14}{36} \times \frac{16}{35} \times 2$ |
| $16 \times 14\ /\ ^{36}C_2 = \frac{16}{45}$ AG | A1 **4** | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Marginal C probs: $\frac{11}{30}$, $\frac{22}{45}$, $\frac{13}{90}$ | B1 | AEF |
| $E(C) = \frac{22}{45} + \frac{26}{90} = \frac{35}{45} = \frac{7}{9}$ | M1 | |
| | A1 **3** | |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| EITHER: $2 \times \frac{1}{42} + \frac{2}{15} + \frac{16}{105}$ | M1 | Other: 0, 1, 2 M1 |
| $E(L) = \frac{8}{9}$, $E(O) = 2 - \frac{15}{9}$ | A1 | $p$: $\frac{29}{42}$, $\frac{2}{7}$, $\frac{1}{42}$ A1 |
| $= \frac{1}{3}$ | A1 **3** | $E(O) = \frac{2}{7} + \frac{2}{42} = \frac{1}{3}$ A1 |
### Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| EITHER: Argument | B2 | e.g. The more $L$s the fewer $C$s |
| OR: Use idea that for independence $P(L \cap C) = P(L)P(C)$ | M1A1 | OR Use conditional probability |
| Conclude that covariance is non-zero | B1 **3** | OR $\text{Cov}(L,C) = -\frac{136}{405}$ M1A1; $L$,$C$ not indep B1 |
| | **[13]** | |
---6 A City Council comprises 16 Labour members, 14 Conservative members and 6 members of Other parties. A sample of two members was chosen at random to represent the Council at an event. The number of Labour members and the number of Conservative members in this sample are denoted by $L$ and $C$ respectively. The joint probability distribution of $L$ and $C$ is given in the following table.
$C$\begin{tabular}{ | c | c c c | }
\multicolumn{4}{c}{$L$} \\
\hline
& 0 & 1 & 2 \\
\hline
0 & $\frac { 1 } { 42 }$ & $\frac { 16 } { 105 }$ & $\frac { 4 } { 21 }$ \\
1 & $\frac { 2 } { 15 }$ & $\frac { 16 } { 45 }$ & 0 \\
2 & $\frac { 13 } { 90 }$ & 0 & 0 \\
\hline
\end{tabular}
(i) Verify the two non-zero probabilities in the table for which $C = 1$.\\
(ii) Find the expected number of Conservatives in the sample.\\
(iii) Find the expected number of Other members in the sample.\\
(iv) Explain why $L$ and $C$ are not independent, and state what can be deduced about $\operatorname { Cov } ( L , C )$.
\hfill \mbox{\textit{OCR S4 2011 Q6 [13]}}