| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Calculate combined outcome probability |
| Difficulty | Moderate -0.3 This is a straightforward tree diagram question testing basic probability rules (multiplication along branches, addition across outcomes) and conditional probability. Part (iv) requires P(A|B) = P(A∩B)/P(B), which is standard S1 content. The calculations are routine with no conceptual challenges, making it slightly easier than average but not trivial due to the multi-part structure and conditional probability component. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Tree diagram with correct shape and labels | B1 | Correct shape and labels |
| Probabilities: S branch 0.4, NS branch 0.6; subsequent branches 0.8/0.2 and 0.8/0.2; final branches 0.3/0.7 | B1 [2] | Correct probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\ (28/125)\) | M1 A1 [2] | Multiplying 3 probs once and 0.7 seen; correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(S, NS, S) + P(NS, S, S) + 0.224\) | M1 B1 | Summing three different 3-factor terms; correct expression for \(P(S, NS, S)\) or \(P(NS, S, S)\) |
| \(= 0.392\ (49/125)\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{stops at first light} \mid \text{stops at exactly 2 lights})\) | ||
| \(= P\frac{(S, NS, S) \text{ or } (S, S, NS)}{0.392}\) | M1 | Summing two 3-factor terms in numerator (need not be different), must be a division |
| \(= \frac{0.4 \times 0.2 \times 0.3 + 0.4 \times 0.8 \times 0.7}{0.392}\) | M1* dep | Dividing by their (iii) if their (iii) \(< 1\), dep on previous M |
| \(= 0.633\ (31/49)\) | A1ft [3] | ft their \(E(X)\) provided \(2 < E(X) < 12\) |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Tree diagram with correct shape and labels | B1 | Correct shape and labels |
| Probabilities: S branch 0.4, NS branch 0.6; subsequent branches 0.8/0.2 and 0.8/0.2; final branches 0.3/0.7 | B1 **[2]** | Correct probabilities |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\ (28/125)$ | M1 A1 **[2]** | Multiplying 3 probs once and 0.7 seen; correct answer |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(S, NS, S) + P(NS, S, S) + 0.224$ | M1 B1 | Summing three different 3-factor terms; correct expression for $P(S, NS, S)$ or $P(NS, S, S)$ |
| $= 0.392\ (49/125)$ | A1 **[3]** | Correct answer |
### Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{stops at first light} \mid \text{stops at exactly 2 lights})$ | | |
| $= P\frac{(S, NS, S) \text{ or } (S, S, NS)}{0.392}$ | M1 | Summing two 3-factor terms in numerator (need not be different), must be a division |
| $= \frac{0.4 \times 0.2 \times 0.3 + 0.4 \times 0.8 \times 0.7}{0.392}$ | M1* dep | Dividing by their (iii) if their (iii) $< 1$, dep on previous M |
| $= 0.633\ (31/49)$ | A1ft **[3]** | ft their $E(X)$ provided $2 < E(X) < 12$ |6 There are three sets of traffic lights on Karinne's journey to work. The independent probabilities that Karinne has to stop at the first, second and third set of lights are $0.4,0.8$ and 0.3 respectively.\\
(i) Draw a tree diagram to show this information.\\
(ii) Find the probability that Karinne has to stop at each of the first two sets of lights but does not have to stop at the third set.\\
(iii) Find the probability that Karinne has to stop at exactly two of the three sets of lights.\\
(iv) Find the probability that Karinne has to stop at the first set of lights, given that she has to stop at exactly two sets of lights.
\hfill \mbox{\textit{CAIE S1 2008 Q6 [10]}}