| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Easy -1.2 This is a straightforward tree diagram question requiring basic probability rules (multiplication along branches, addition across paths) and simple conditional probability using P(A|B) = P(A∩B)/P(B). All values are given directly, requiring only routine calculation with no problem-solving insight or complex manipulation. |
| 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 for method of transport with correct probabilities (P: 0.35, B: 0.44, C: 0.21) | B1 | |
| Fully correct labelled branches with correct probabilities for lateness (P→Y: 0.3, N: 0.7; B→Y: 0.8, N: 0.2; C→Y: 0, N: 1) | B1 | Either 1 branch after W or 2 branches with prob 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.35 \times 0.3 + 0.44 \times 0.8 \ (+0)\) | M1 | |
| \(0.457\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{not B} \mid \text{not fruit}) = \frac{P(B' \cap F')}{P(F')}\) | M1 | |
| \(\frac{0.35\times0.7 + 0.21\times1}{1 - their(\mathbf{b})}\) | M1 | |
| \(\frac{0.455}{0.543}\) | M1 | M1 for \(1 - their(\mathbf{b})\) or summing three appropriate 2-factor probabilities, correct or consistent with their tree diagram as denominator |
| \(0.838\) or \(\frac{455}{543}\) | A1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Fully correct labelled tree for method of transport with correct probabilities (P: 0.35, B: 0.44, C: 0.21) | B1 | |
| Fully correct labelled branches with correct probabilities for lateness (P→Y: 0.3, N: 0.7; B→Y: 0.8, N: 0.2; C→Y: 0, N: 1) | B1 | Either 1 branch after W or 2 branches with prob 0 |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.35 \times 0.3 + 0.44 \times 0.8 \ (+0)$ | M1 | |
| $0.457$ | A1 | |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{not B} \mid \text{not fruit}) = \frac{P(B' \cap F')}{P(F')}$ | M1 | |
| $\frac{0.35\times0.7 + 0.21\times1}{1 - their(\mathbf{b})}$ | M1 | |
| $\frac{0.455}{0.543}$ | M1 | M1 for $1 - their(\mathbf{b})$ or summing three appropriate 2-factor probabilities, correct or consistent with their tree diagram as denominator |
| $0.838$ or $\frac{455}{543}$ | A1 | |5 On Mondays, Rani cooks her evening meal. She has a pizza, a burger or a curry with probabilities $0.35,0.44,0.21$ respectively. When she cooks a pizza, Rani has some fruit with probability 0.3 . When she cooks a burger, she has some fruit with probability 0.8 . When she cooks a curry, she never has any fruit.
\begin{enumerate}[label=(\alph*)]
\item Draw a fully labelled tree diagram to represent this information.
\item Find the probability that Rani has some fruit.
\item Find the probability that Rani does not have a burger given that she does not have any fruit.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q5 [8]}}