| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Sequential selection without replacement |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram question with sequential selection without replacement. Part (i) requires basic conditional probability calculations, part (ii) is a 'show that' requiring summing two branches (routine verification), and part (iii) applies standard formulas for expectation and variance given a complete probability distribution. All steps are mechanical applications of standard S1 techniques with no problem-solving insight required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 6 }\) | \(\frac { 3 } { 5 }\) | \(\frac { 7 } { 30 }\) |
5 A bag contains 4 blue discs and 6 red discs. Chloe takes a disc from the bag. If this disc is red, she takes 2 more discs. If not, she takes 1 more disc. Each disc is taken at random and no discs are replaced.\\
(i) Complete the probability tree diagram in your Answer Book, showing all the probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{48ffcd44-d933-40e0-818a-20d6db607298-4_730_1203_529_511}
The total number of blue discs that Chloe takes is denoted by $X$.\\
(ii) Show that $\mathrm { P } ( X = 1 ) = \frac { 3 } { 5 }$.
The complete probability distribution of $X$ is given below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 6 }$ & $\frac { 3 } { 5 }$ & $\frac { 7 } { 30 }$ \\
\hline
\end{tabular}
\end{center}
(iii) Calculate $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR S1 2011 Q5 [9]}}