| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Conditional probability tree diagram |
| Difficulty | Standard +0.3 This is a conditional probability problem with a tree diagram that requires careful tracking of states (who won the previous set) and calculating probabilities along branches. While it involves multiple steps and conditional probabilities, the structure is straightforward once the tree is drawn, and the calculations are routine multiplication and addition of probabilities. It's slightly easier than average because tree diagrams are a standard S1 technique and the problem clearly guides students through the setup. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | 0.7 Jimmy |
| Answer | Marks |
|---|---|
| . 6 Alan | G1 |
| Answer | Marks |
|---|---|
| [3] | Do a vertical scan and |
| Answer | Marks |
|---|---|
| Final column | All indep |
| Answer | Marks |
|---|---|
| (ii) | P(Alan wins) |
| = (0.4×0.3×0.6) + (0.6×0.4×0.3) + (0.6×0.6) = 0.504 | M1 |
| Answer | Marks |
|---|---|
| [3] | For any one ‘correct’ |
| Answer | Marks |
|---|---|
| CAO | FT their tree for both M marks |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) | P(Ends after 4) = (0.4×0.7) + (0.6×0.6) = 0.28 + 0.36 = 0.64 | M1 |
| Answer | Marks |
|---|---|
| [2] | For both products |
| CAO | FT their tree for M mark but not for A mark |
Question 2:
2 | (i) | 0.7 Jimmy
0.4 Jimmy
Jimmy
Alan
0.3
Alan
.4 0.6
0.7 Jimmy
.6 Jimmy
. 4
Alan 0.3 Alan
. 6 Alan | G1
G1
G1
[3] | Do a vertical scan and
give:
First column
Second column
Final column | All indep
All probs must be correct
Without extra branches in final column
Ignore anything before third set
Allow labels ‘win’ and ‘lose’ in place of
Jimmy and Alan respectively but if no
labels, no marks
(ii) | P(Alan wins)
= (0.4×0.3×0.6) + (0.6×0.4×0.3) + (0.6×0.6) = 0.504 | M1
M1
A1
[3] | For any one ‘correct’
product
For all three ‘correct’
products and no extras
CAO | FT their tree for both M marks
Provided correct number of terms in
product(s) for both M1’s
(iii) | P(Ends after 4) = (0.4×0.7) + (0.6×0.6) = 0.28 + 0.36 = 0.64 | M1
A1
[2] | For both products
CAO | FT their tree for M mark but not for A mark
Provided two terms in each productJimmy and Alan are playing a tennis match against each other. The winner of the match is the first player to win three sets. Jimmy won the first set and Alan won the second set. For each of the remaining sets, the probability that Jimmy wins a set is
• 0.7 if he won the previous set,
• 0.4 if Alan won the previous set.
It is not possible to draw a set.
\begin{enumerate}[label=(\roman*)]
\item Draw a probability tree diagram to illustrate the possible outcomes for each of the remaining sets. [3]
\item Find the probability that Alan wins the match. [3]
\item Find the probability that the match ends after exactly four sets have been played. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q2 [8]}}