| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Calculate combined outcome probability |
| Difficulty | Moderate -0.8 This is a straightforward probability tree question with clearly stated probabilities and standard calculations (total probability and conditional probability). The tree has only 2-3 levels with simple multiplication and addition of given probabilities. Part (iii) requires Bayes' theorem but in a routine application. Easier than average A-level as it's purely mechanical with no problem-solving insight needed. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tree diagram showing:<br>First branch: Accept (0.2), Retest (0.3), Reject (0.5)<br>Second branches from Retest: Accept (0.2), Retest (0.5), Reject (0.3)<br>Third branches from second Retest: Accept (0.4), Retest (0.3), Reject (0.6) | Alternative version of tree diagram for Q2(i) | |
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Do a vertical scan and give: First column | G1 | Allow labels such as A, R, F(Fail) etc. All probabilities correct |
| Second column | G1 | All probabilities correct |
| Final column | G1 | All probabilities correct |
| Do not award if first two branches missing. Branches two and three should come out of 'retest'. If any labels missing or incorrect allow max 2/3. Do not allow misreads here as all FT (eg 0.3 and 0.5 reversed) | ||
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Accepted}) = 0.2 + (0.3 \times 0.2) + (0.3 \times 0.3 \times 0.4)\) | M1 | For second or third product |
| \(= 0.2 + 0.06 + 0.036 = 0.296\) | A1 | CAO |
| Total | [2] | FT their tree provided correct numbers of terms and correct structure of 3, 3, 2 branches. Allow 37/125 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{At least one retest given accepted}) = \frac{P(\text{At least one retest and accepted})}{P(\text{Accepted})}\) | M1 | For numerator. FT their tree provided correct numbers of terms and correct structure of 3, 3, 2 branches. for both M1's |
| \(= \frac{0.3 \times 0.2 + 0.3 \times 0.3 \times 0.4}{0.296} = \frac{0.096}{0.296}\) | M1 | For denominator |
| \(= 0.324\) | A1 | FT their 0.296 and 0.096. Allow 12/37 oe |
| Total | [3] |
## (i)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| Tree diagram showing:<br>First branch: Accept (0.2), Retest (0.3), Reject (0.5)<br>Second branches from Retest: Accept (0.2), Retest (0.5), Reject (0.3)<br>Third branches from second Retest: Accept (0.4), Retest (0.3), Reject (0.6) | | Alternative version of tree diagram for Q2(i) |
| **Total** | **[3]** | |
## (i) (Detailed)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| Do a vertical scan and give: First column | G1 | Allow labels such as A, R, F(Fail) etc. All probabilities correct |
| Second column | G1 | All probabilities correct |
| Final column | G1 | All probabilities correct |
| | | Do not award if first two branches missing. Branches two and three should come out of 'retest'. If any labels missing or incorrect allow max 2/3. Do not allow misreads here as all FT (eg 0.3 and 0.5 reversed) |
| **Total** | **[3]** | |
## (ii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $P(\text{Accepted}) = 0.2 + (0.3 \times 0.2) + (0.3 \times 0.3 \times 0.4)$ | M1 | For second or third product |
| $= 0.2 + 0.06 + 0.036 = 0.296$ | A1 | CAO |
| **Total** | **[2]** | FT their tree provided correct numbers of terms and correct structure of 3, 3, 2 branches. Allow 37/125 oe |
## (iii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $P(\text{At least one retest given accepted}) = \frac{P(\text{At least one retest and accepted})}{P(\text{Accepted})}$ | M1 | For numerator. FT their tree provided correct numbers of terms and correct structure of 3, 3, 2 branches. for both M1's |
| $= \frac{0.3 \times 0.2 + 0.3 \times 0.3 \times 0.4}{0.296} = \frac{0.096}{0.296}$ | M1 | For denominator |
| $= 0.324$ | A1 | FT their 0.296 and 0.096. Allow 12/37 oe |
| **Total** | **[3]** | |
---Candidates applying for jobs in a large company take an aptitude test, as a result of which they are either accepted, rejected or retested, with probabilities 0.2, 0.5 and 0.3 respectively. When a candidate is retested for the first time, the three possible outcomes and their probabilities remain the same as for the original test. When a candidate is retested for the second time there are just two possible outcomes, accepted or rejected, with probabilities 0.4 and 0.6 respectively.
\begin{enumerate}[label=(\roman*)]
\item Draw a probability tree diagram to illustrate the outcomes. [3]
\item Find the probability that a randomly selected candidate is accepted. [2]
\item Find the probability that a randomly selected candidate is retested at least once, given that this candidate is accepted. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2014 Q2 [8]}}