| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| 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 S4 question requiring verification of a joint probability entry using basic counting, then computing covariance using the standard formula. The joint distribution is given, so students only need to apply memorized formulas for E(F), E(S), and E(FS). While it's Further Maths content, it's a routine textbook exercise with no conceptual challenges. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(S\) | |||
| 0 | 1 | 2 | |
| 0 | \(\frac { 1 } { 8 }\) | \(\frac { 1 } { 8 }\) | 0 |
| 1 | \(\frac { 1 } { 8 }\) | \(\frac { 1 } { 4 }\) | \(\frac { 1 } { 8 }\) |
| 2 | 0 | \(\frac { 1 } { 8 }\) | \(\frac { 1 } { 8 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F=1, S=1\) requires HTH or THT | M1 | Clear method – not just multiplication of probs |
| Probability \(= \frac{1}{8} + \frac{1}{8} = \frac{1}{4}\) AG | A1 | \(SC\ \frac{2}{8} = \frac{1}{4}\) ONLY seen B1. NOT \(\frac{1}{2} \times \frac{1}{2}\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Marginals: 0, 1, 2; \(p(S)\): \(\frac{1}{4}, \frac{1}{2}, \frac{1}{4}\); \(p(F)\): \(\frac{1}{4}, \frac{1}{2}, \frac{1}{4}\) | M1, A1 | Correct method, can be implied by (i). Both correct, can be implied by e.g. \(E(S)=E(F)=1\) or symmetry |
| \(E(S) = 1\times\frac{1}{2} + 2\times\frac{1}{4} = 1 = E(F)\) | B1 | |
| \(E(SF) = \frac{1}{4} + \frac{2}{8} + \frac{2}{8} + \frac{4}{8} = 1\frac{1}{4}\) | M1* | |
| \(\text{Cov}(S,F) = E(SF) - E(S)E(F)\) | *M1 | |
| \(= \frac{1}{4}\) | A1 | |
| [6] |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F=1, S=1$ requires HTH or THT | M1 | Clear method – not just multiplication of probs |
| Probability $= \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$ AG | A1 | $SC\ \frac{2}{8} = \frac{1}{4}$ ONLY seen B1. NOT $\frac{1}{2} \times \frac{1}{2}$ |
| **[2]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Marginals: 0, 1, 2; $p(S)$: $\frac{1}{4}, \frac{1}{2}, \frac{1}{4}$; $p(F)$: $\frac{1}{4}, \frac{1}{2}, \frac{1}{4}$ | M1, A1 | Correct method, can be implied by (i). Both correct, can be implied by e.g. $E(S)=E(F)=1$ or symmetry |
| $E(S) = 1\times\frac{1}{2} + 2\times\frac{1}{4} = 1 = E(F)$ | B1 | |
| $E(SF) = \frac{1}{4} + \frac{2}{8} + \frac{2}{8} + \frac{4}{8} = 1\frac{1}{4}$ | M1* | |
| $\text{Cov}(S,F) = E(SF) - E(S)E(F)$ | *M1 | |
| $= \frac{1}{4}$ | A1 | |
| **[6]** | | |
---1
\begin{center}
\begin{tabular}{ c | c | c c c | }
& \multicolumn{3}{c}{$S$} \\
\hline
& 0 & 1 & 2 \\
\hline
0 & $\frac { 1 } { 8 }$ & $\frac { 1 } { 8 }$ & 0 \\
1 & $\frac { 1 } { 8 }$ & $\frac { 1 } { 4 }$ & $\frac { 1 } { 8 }$ \\
2 & 0 & $\frac { 1 } { 8 }$ & $\frac { 1 } { 8 }$ \\
\hline
\end{tabular}
\end{center}
An unbiased coin is tossed three times. The random variables $F$ and $S$ denote the total number of heads that occur in the first two tosses and the total number of heads that occur in the last two tosses respectively. The table above shows the joint probability distribution of $F$ and $S$.\\
(i) Show how the entry $\frac { 1 } { 4 }$ in the table is obtained.\\
(ii) Find $\operatorname { Cov } ( F , S )$.
\hfill \mbox{\textit{OCR S4 2013 Q1 [8]}}