| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Tree diagram with two-stage events |
| Difficulty | Easy -1.2 This is a straightforward two-stage tree diagram problem with standard conditional probability calculation using Bayes' theorem. Part (a) requires only organizing given probabilities into a tree structure, and part (b) is a direct application of P(A|B) = P(A∩B)/P(B) with simple arithmetic—no conceptual challenges or problem-solving insight needed. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Fully correct labelled tree diagram for method of transport (C, B, W with probabilities 0.2, 0.45, 0.35) with correct probabilities | B1 | Must show all three branches with correct labels and probabilities |
| Fully correct labelled branches for lateness (E, NE) with correct probabilities: C→(0.6, 0.4), B→(0.9, 0.1), W→(1, 0) | B1 | Either 1 branch after W or 2 branches with probability 0 |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(C\ | E) = \dfrac{P(C \cap E)}{P(E)} = \dfrac{0.2 \times 0.6}{0.2 \times 0.6 + 0.45 \times 0.1 + 0.35 \times 1}\) | M1 |
| Summing three appropriate 2-factor probabilities in denominator | M1 | |
| \(\dfrac{0.12}{0.515}\) | A1 | Correct numerator and denominator |
| \(0.233\) or \(\dfrac{12}{515}\) | A1 | |
| Total: 4 |
## Question 1:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Fully correct labelled tree diagram for method of transport (C, B, W with probabilities 0.2, 0.45, 0.35) with correct probabilities | **B1** | Must show all three branches with correct labels and probabilities |
| Fully correct labelled branches for lateness (E, NE) with correct probabilities: C→(0.6, 0.4), B→(0.9, 0.1), W→(1, 0) | **B1** | Either 1 branch after W or 2 branches with probability 0 |
| | **Total: 2** | |
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(C\|E) = \dfrac{P(C \cap E)}{P(E)} = \dfrac{0.2 \times 0.6}{0.2 \times 0.6 + 0.45 \times 0.1 + 0.35 \times 1}$ | **M1** | Correct conditional probability structure seen |
| Summing three appropriate 2-factor probabilities in denominator | **M1** | |
| $\dfrac{0.12}{0.515}$ | **A1** | Correct numerator and denominator |
| $0.233$ or $\dfrac{12}{515}$ | **A1** | |
| | **Total: 4** | |1 Juan goes to college each day by any one of car or bus or walking. The probability that he goes by car is 0.2 , the probability that he goes by bus is 0.45 and the probability that he walks is 0.35 . When Juan goes by car, the probability that he arrives early is 0.6 . When he goes by bus, the probability that he arrives early is 0.1 . When he walks he always arrives early.
\begin{enumerate}[label=(\alph*)]
\item Draw a fully labelled tree diagram to represent this information.
\item Find the probability that Juan goes to college by car given that he arrives early.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q1 [6]}}