| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Bayes with complementary outcome |
| Difficulty | Moderate -0.3 This is a standard two-stage tree diagram problem with straightforward Bayes' theorem application. Students must find P(social media)=0.5 by complement, construct the tree, calculate P(no reply) using the law of total probability, then apply Bayes' theorem. While it requires multiple steps and careful organization, it follows a completely standard template for AS-level conditional probability with no novel insight required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Fully correct labelled tree with correct probabilities for 'Send' branches | B1 | Fully correct labelled tree with correct probabilities for 'Send' |
| Fully correct labelled branches with correct probabilities for 'reply' | B1 | Fully correct labelled branches with correct probabilities for the 'reply' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{email}\ | NR) = \dfrac{P(\text{email} \cap NR)}{P(NR)} = \dfrac{0.2 \times 0.85}{0.3 \times 0.6 + 0.2 \times 0.85 + 0.5 \times 0.4}\) | M1 |
| \(= \dfrac{0.17}{0.18 + 0.17 + 0.2} = \dfrac{0.17}{0.55}\) | M1 | Summing three appropriate 2-factor probabilities, consistent with their tree diagram, seen anywhere; 0.55 oe (can be unsimplified) seen as denom of a fraction |
| \(= 0.309,\ \dfrac{17}{55}\) | A1 | |
| Correct answer | A1 | Correct answer |
## Question 2(i):
Tree diagram with three branches (text: 0.3, email: 0.2, social media: 0.5) and reply branches (R/NR):
- Text: R = 0.4, NR = 0.6
- Email: R = 0.15, NR = 0.85
- Social media: R = 0.6, NR = 0.4
| Answer | Marks | Guidance |
|--------|-------|----------|
| Fully correct labelled tree with correct probabilities for 'Send' branches | B1 | Fully correct labelled tree with correct probabilities for 'Send' |
| Fully correct labelled branches with correct probabilities for 'reply' | B1 | Fully correct labelled branches with correct probabilities for the 'reply' |
---
## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{email}\|NR) = \dfrac{P(\text{email} \cap NR)}{P(NR)} = \dfrac{0.2 \times 0.85}{0.3 \times 0.6 + 0.2 \times 0.85 + 0.5 \times 0.4}$ | M1 | $P(\text{email}) \times P(NR)$ seen as numerator of a fraction, consistent with their tree diagram |
| $= \dfrac{0.17}{0.18 + 0.17 + 0.2} = \dfrac{0.17}{0.55}$ | M1 | Summing three appropriate 2-factor probabilities, consistent with their tree diagram, seen anywhere; 0.55 oe (can be unsimplified) seen as denom of a fraction |
| $= 0.309,\ \dfrac{17}{55}$ | A1 | |
| Correct answer | A1 | Correct answer |
---2 Megan sends messages to her friends in one of 3 different ways: text, email or social media. For each message, the probability that she uses text is 0.3 and the probability that she uses email is 0.2 . She receives an immediate reply from a text message with probability 0.4 , from an email with probability 0.15 and from social media with probability 0.6 .\\
(i) Draw a fully labelled tree diagram to represent this information.\\
(ii) Given that Megan does not receive an immediate reply to a message, find the probability that the message was an email.\\
\hfill \mbox{\textit{CAIE S1 2019 Q2 [6]}}