| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 12 |
| Topic | Discrete Probability Distributions |
| Type | Sum or product of two independent values |
| Difficulty | Standard +0.3 This is a multi-part Stats 1 question involving discrete probability distributions with independent dice. While it requires careful bookkeeping (sample space diagram, conditional probability, identifying mutually exclusive events), the techniques are all standard for this module. Part (ii) tests understanding of equally likely outcomes, and parts (iii)-(v) apply routine probability formulas. The complexity comes from multiple parts rather than conceptual difficulty, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.04a Discrete probability distributions |
| Biased die | ||||||
| \(y\) | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm { P } ( Y = y )\) | 0.3 | 0.25 | 0.2 | 0.14 | 0.1 | 0.01 |
| Answer | Marks | Guidance |
|---|---|---|
| Biased die, Y | ||
| 1 | 2 | 3 |
| 1 | 3 | 5 |
| 2 | 4 | 6 |
| Fair die, X | 3 | 5 |
| 4 | 5 | 6 |
| 5 | 6 | 7 |
| 6 | 7 | 8 |
| B2 | 1.1a, 1.1 | B1 for \(\geq 30\) values correct |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Four 10s circled or otherwise indicated | B1 | Must be exactly four 10s in table. |
| [1] |
| Answer | Marks |
|---|---|
| Outcomes not equally likely oe | E1 |
| [1] |
| Answer | Marks |
|---|---|
| \(= \frac{13}{200}\) or \(0.065\) | M1, A1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{6} \times 0.14 + \frac{1}{6} \times 0.14 + \frac{1}{6} \times 0.01\) | M1 | \((= \frac{29}{600}\) or \(0.0483(3 \text{ sf}))\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{\text{P}(S = 10 \text{ & one score} = 4)}{\text{P}(S = 10)} = \frac{29}{600} \div \frac{13}{200}\) | M1 | ft their (iii), dep 1st M1 gained in (iv) |
| \(= \frac{29}{39}\) or \(0.744\) (3 sf) | A1 | cao |
| [3] |
| Answer | Marks |
|---|---|
| \(n = 15\) | B1 |
| [1] |
## 12(i)(a)
| | Biased die, Y | | | | | |
|---|---|---|---|---|---|---|
| | **1** | **2** | **3** | **4** | **5** | **6** |
| **1** | 3 | 5 | 7 | 9 | 11 | 13 |
| **2** | 4 | 6 | 8 | 10 | 12 | 14 |
| Fair die, X | **3** | 5 | 7 | 9 | 11 | 13 | 15 |
| **4** | 5 | 6 | 7 | 8 | 9 | 10 |
| **5** | 6 | 7 | 8 | 9 | 10 | 11 |
| **6** | 7 | 8 | 9 | 10 | 11 | 12 |
| B2 | 1.1a, 1.1 | B1 for $\geq 30$ values correct
| [2]
## 12(i)(b)
Four 10s circled or otherwise indicated | B1 | Must be exactly four 10s in table.
| [1]
## 12(ii)
Outcomes not equally likely oe | E1 |
| [1]
## 12(iii)
$\frac{1}{6} \times 0.14 + \frac{1}{6} \times 0.14 + \frac{1}{6} \times 0.1 + \frac{1}{6} \times 0.01$
$= \frac{13}{200}$ or $0.065$ | M1, A1 |
| [2]
## 12(iv)
$\text{P}(S = 10 \text{ & one score} = 4) =$
$\frac{1}{6} \times 0.14 + \frac{1}{6} \times 0.14 + \frac{1}{6} \times 0.01$ | M1 | $(= \frac{29}{600}$ or $0.0483(3 \text{ sf}))$
$\text{P}(\text{One score} = 4 \mid S = 10)$
$= \frac{\text{P}(S = 10 \text{ & one score} = 4)}{\text{P}(S = 10)} = \frac{29}{600} \div \frac{13}{200}$ | M1 | ft their (iii), dep 1st M1 gained in (iv)
$= \frac{29}{39}$ or $0.744$ (3 sf) | A1 | cao
| [3]
## 12(v)
$n = 15$ | B1 |
| [1]12 Rob has two six-sided dice, each with sides numbered 1, 2, 3, 4, 5, 6.\\
One dice is fair. The other dice is biased, with probabilities as shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
\multicolumn{7}{|c|}{Biased die} \\
\hline
$y$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( Y = y )$ & 0.3 & 0.25 & 0.2 & 0.14 & 0.1 & 0.01 \\
\hline
\end{tabular}
\end{center}
Rob throws each dice once and notes the two scores, $X$ on the fair dice and $Y$ on the biased dice. He then calculates the value of the variable $S$ which is defined as follows.
\begin{itemize}
\item If $X \leqslant 3$, then $S = X + 2 Y$.
\item If $X > 3$, then $S = X + Y$.
\begin{enumerate}[label=(\roman*)]
\item (a) Draw up a sample space diagram showing all the possible outcomes and the corresponding values of $S$.\\
(b) On your diagram, circle the four cells where the value $S = 10$ occurs.
\item Explain the mistake in the following calculation.
\end{itemize}
$$\mathrm { P } ( S = 10 ) = \frac { \text { Number of outcomes giving } S = 10 } { \text { Total number of outcomes } } = \frac { 4 } { 36 } = \frac { 1 } { 9 } .$$
\item Find the correct value of $\mathrm { P } ( S = 10 )$.
\item Given that $S = 10$, find the probability that the score on one of the dice is 4 .
\item The events " $X = 1$ or 2 " and " $S = n$ " are mutually exclusive. Given that $\mathrm { P } ( S = n ) \neq 0$, find the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2018 Q12 [12]}}