| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Bayes with complementary outcome |
| Difficulty | Moderate -0.5 This is a straightforward application of the law of total probability and Bayes' theorem with clearly stated probabilities and a simple two-stage tree diagram structure. Part (i) requires basic probability multiplication and addition, while part (ii) is a standard Bayes' theorem calculation using complementary probabilities. The question is slightly easier than average because it involves only two branches, clearly labeled probabilities, and follows a standard textbook format with no conceptual tricks or multi-step reasoning required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.6 \times 0.2 + 0.4 \times 0.32\) | M1 | Addition of 2 two-factor terms \(0.6 \times a + 0.4 \times b\) |
| \(= 0.248, \dfrac{31}{125}\) | A1 | CAO |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(GS \mid \text{Not Red socks}) = \dfrac{0.4 \times 0.68}{1-(i)}\) | B1 | Correct [unsimplified] numerator seen in fraction |
| M1 | \(1 -\) their (i) as denominator in fraction | |
| \(= 0.362, \dfrac{17}{47}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(GS \mid \text{Not Red socks}) = \dfrac{0.4 \times 0.68}{0.6 \times 0.8 + 0.4 \times 0.68}\) | B1 | Correct [unsimplified] numerator seen in fraction |
| M1 | Correct or (their (i))' as denominator in fraction | |
| \(= 0.362, \dfrac{17}{47}\) | A1 | |
| Total: 3 |
## Question 1:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.6 \times 0.2 + 0.4 \times 0.32$ | M1 | Addition of 2 two-factor terms $0.6 \times a + 0.4 \times b$ |
| $= 0.248, \dfrac{31}{125}$ | A1 | CAO |
| | **Total: 2** | |
---
### Part (ii):
**Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(GS \mid \text{Not Red socks}) = \dfrac{0.4 \times 0.68}{1-(i)}$ | B1 | Correct [unsimplified] numerator seen in fraction |
| | M1 | $1 -$ their **(i)** as denominator in fraction |
| $= 0.362, \dfrac{17}{47}$ | A1 | |
**Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(GS \mid \text{Not Red socks}) = \dfrac{0.4 \times 0.68}{0.6 \times 0.8 + 0.4 \times 0.68}$ | B1 | Correct [unsimplified] numerator seen in fraction |
| | M1 | Correct or (their **(i)**)' as denominator in fraction |
| $= 0.362, \dfrac{17}{47}$ | A1 | |
| | **Total: 3** | |1 On each day that Tamar goes to work, he wears either a blue suit with probability 0.6 or a grey suit with probability 0.4 . If he wears a blue suit then the probability that he wears red socks is 0.2 . If he wears a grey suit then the probability that he wears red socks is 0.32 .\\
(i) Find the probability that Tamar wears red socks on any particular day that he is at work.\\
(ii) Given that Tamar is not wearing red socks at work, find the probability that he is wearing a grey suit.\\
\hfill \mbox{\textit{CAIE S1 2019 Q1 [5]}}