| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | December |
| Marks | 11 |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Standard +0.8 This is a conditional probability problem requiring careful construction of a tree diagram with changing probabilities (sampling without replacement), calculation of specific outcomes across multiple branches, and application of Bayes' theorem for part (c). The problem demands systematic case analysis and is significantly more challenging than routine Stats 1 questions, though not requiring advanced techniques beyond the syllabus. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space |
Mr Jones has 3 tins of beans and 2 tins of pears. His daughter has removed the labels for a school project, and the tins are identical in appearance. Mr Jones opens tins in turn until he has opened at least 1 tin of beans and at least 1 tin of pears. He does not open any remaining tins.
\begin{enumerate}[label=(\alph*)]
\item Draw a tree diagram to illustrate this situation, labelling each branch with its associated probability. [3]
\item Find the probability that Mr Jones opens exactly 3 tins. [3]
\item It is given that the last tin Mr Jones opens is a tin of pears. Find the probability that he opens exactly 3 tins. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2018 Q14 [11]}}