| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Moderate -0.8 This is a straightforward two-stage tree diagram problem with basic conditional probability using Bayes' theorem. The calculations involve simple multiplication and division of given probabilities with no conceptual challenges—standard S1 material that's easier than average A-level questions. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tree diagram with correct shape: first branches \(0.85\) (Pass), \(0.15\) (Fail); second branches from Pass: \(0.65\) (Pass), Fail; from Fail: \(0.35\) (Fail) | M1 | Correct shape |
| All correct labels and probabilities | A1 | All correct labels and probabilities |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(F\mid P) = \frac{P(F\cap P)}{P(P)}\) | M1 | \(P(P)\) consistent with their tree diagram seen anywhere |
| \(= \frac{0.15\times0.65}{0.85+0.15\times0.65}\) or \(\frac{0.15\times0.65}{1-0.15\times0.35}\) | A1 | Correct unsimplified \(P(P)\) seen as numerator or denominator of a fraction |
| \(= \frac{0.0975}{0.9475}\) | M1 | \(P(F\cap P)\) found as correct product or consistent with their tree diagram seen as numerator or denominator of a fraction |
| \(= \frac{39}{379} = 0.103\) | A1 | |
| Total: 4 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tree diagram with correct shape: first branches $0.85$ (Pass), $0.15$ (Fail); second branches from Pass: $0.65$ (Pass), Fail; from Fail: $0.35$ (Fail) | M1 | Correct shape |
| All correct labels and probabilities | A1 | All correct labels and probabilities |
| **Total: 2** | | |
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## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(F\mid P) = \frac{P(F\cap P)}{P(P)}$ | M1 | $P(P)$ consistent with their tree diagram seen anywhere |
| $= \frac{0.15\times0.65}{0.85+0.15\times0.65}$ or $\frac{0.15\times0.65}{1-0.15\times0.35}$ | A1 | Correct unsimplified $P(P)$ seen as numerator or denominator of a fraction |
| $= \frac{0.0975}{0.9475}$ | M1 | $P(F\cap P)$ found as correct product or consistent with their tree diagram seen as numerator or denominator of a fraction |
| $= \frac{39}{379} = 0.103$ | A1 | |
| **Total: 4** | | |
---3 At the end of a revision course in mathematics, students have to pass a test to gain a certificate. The probability of any student passing the test at the first attempt is 0.85 . Those students who fail are allowed to retake the test once, and the probability of any student passing the retake test is 0.65 .\\
(i) Draw a fully labelled tree diagram to show all the outcomes.\\
(ii) Given that a student gains the certificate, find the probability that this student fails the test on the first attempt.\\
\hfill \mbox{\textit{CAIE S1 2017 Q3 [6]}}