| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Standard +0.8 This multi-stage tree diagram problem requires careful tracking of three rounds with changing probabilities and a stopping condition. Part (iii) involves conditional probability P(retested ≥1 | accepted), requiring identification of multiple acceptance paths and application of Bayes' theorem. While the individual calculations are standard S1 material, the complexity of managing multiple branches and the conditional probability reasoning elevate this above average difficulty. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram with branches: First split 0.5/0.5 (Accept/Reject), from Reject: 0.3 Retest, 0.2 Accept, 0.5 Reject; from Retest: 0.4 Accept, 0.3 Retest, 0.6 Reject | G1, G1, G1 [3] | First column all probabilities correct; Second column all probabilities correct; Final column all probabilities correct; Do not award if first two branches missing; If any labels missing allow max 2/3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Accepted}) = 0.2 + (0.3 \times 0.2) + (0.3 \times 0.3 \times 0.4) = 0.2 + 0.06 + 0.036 = 0.296\) | M1, A1 [2] | M1 for second or third product; CAO; Allow \(\frac{37}{125}\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{At least one retest} \mid \text{accepted}) = \frac{P(\text{At least one retest and accepted})}{P(\text{Accepted})} = \frac{0.3 \times 0.2 + 0.3 \times 0.3 \times 0.4}{0.296} = \frac{0.096}{0.296} = 0.324\) | M1, M1, A1 [3] | M1 for numerator; M1 for denominator; FT their 0.296 and 0.096; Allow 0.32 with working; Allow \(\frac{12}{37}\) or equivalent |
## Question 3:
**(i)**
Tree diagram with branches: First split 0.5/0.5 (Accept/Reject), from Reject: 0.3 Retest, 0.2 Accept, 0.5 Reject; from Retest: 0.4 Accept, 0.3 Retest, 0.6 Reject | G1, G1, G1 [3] | First column all probabilities correct; Second column all probabilities correct; Final column all probabilities correct; Do not award if first two branches missing; If any labels missing allow max 2/3
**(ii)**
$P(\text{Accepted}) = 0.2 + (0.3 \times 0.2) + (0.3 \times 0.3 \times 0.4) = 0.2 + 0.06 + 0.036 = 0.296$ | M1, A1 [2] | M1 for second or third product; CAO; Allow $\frac{37}{125}$ or equivalent
**(iii)**
$P(\text{At least one retest} \mid \text{accepted}) = \frac{P(\text{At least one retest and accepted})}{P(\text{Accepted})} = \frac{0.3 \times 0.2 + 0.3 \times 0.3 \times 0.4}{0.296} = \frac{0.096}{0.296} = 0.324$ | M1, M1, A1 [3] | M1 for numerator; M1 for denominator; FT their 0.296 and 0.096; Allow 0.32 with working; Allow $\frac{12}{37}$ or equivalent
---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.\\
(i) Draw a probability tree diagram to illustrate the outcomes.\\
(ii) Find the probability that a randomly selected candidate is accepted.\\
(iii) Find the probability that a randomly selected candidate is retested at least once, given that this candidate is accepted.
\hfill \mbox{\textit{OCR MEI S1 Q3 [8]}}