| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Calculate end-outcome probability |
| Difficulty | Moderate -0.3 This is a standard two-stage probability problem using tree diagrams with straightforward conditional probability calculations. While it requires multiple steps and understanding of conditional probability (part iii), the setup is clear, the numbers are simple, and all parts follow directly from constructing a basic probability tree—slightly easier than average for A-level. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles5.02a Discrete probability distributions: general |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(W, R) = \frac{1}{6} \times \frac{7}{10} = \frac{7}{60}\ (0.117)\) | M1, A1 [2] | For a single product with 6 and 10 in denoms; Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(R, R) = \frac{5}{6} \times \frac{8}{10} = \frac{40}{60}\) | M1 | For finding their \(P(R,R)\) and adding it to their (i) |
| \(P(\text{red}) = \frac{47}{60}\ (= 0.783)\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(R | R) = \frac{P(R \cap R)}{P(R)}\) | M1 |
| \(= \frac{40}{47}\ (= 0.851)\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(R,W) = \frac{5}{6} \times \frac{2}{10} = \frac{10}{60}\); \(P(W,W) = \frac{1}{6} \times \frac{3}{10} = \frac{3}{60}\) | B1 | \(x = 0, 1, 2\) only seen, no probabilities needed |
| B1 | One correct probability | |
| \(P(X=x)\): \(x=0\): \(3/60\); \(x=1\): \(17/60\); \(x=2\): \(40/60\) | B1, B1ft [4] | Another correct probability; ft if only one prob correct and \(\Sigma p = 1\) |
## Question 7:
**Part (i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(W, R) = \frac{1}{6} \times \frac{7}{10} = \frac{7}{60}\ (0.117)$ | M1, A1 **[2]** | For a single product with 6 and 10 in denoms; Correct answer |
**Part (ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(R, R) = \frac{5}{6} \times \frac{8}{10} = \frac{40}{60}$ | M1 | For finding their $P(R,R)$ and adding it to their (i) |
| $P(\text{red}) = \frac{47}{60}\ (= 0.783)$ | A1 **[2]** | Correct answer |
**Part (iii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(R|R) = \frac{P(R \cap R)}{P(R)}$ | M1 | Their $P(R,R)$ / their $P(R)$ ie something $\times 5/6 \div$ their (ii) |
| $= \frac{40}{47}\ (= 0.851)$ | A1 **[2]** | Correct answer |
**Part (iv)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(R,W) = \frac{5}{6} \times \frac{2}{10} = \frac{10}{60}$; $P(W,W) = \frac{1}{6} \times \frac{3}{10} = \frac{3}{60}$ | B1 | $x = 0, 1, 2$ only seen, no probabilities needed |
| | B1 | One correct probability |
| $P(X=x)$: $x=0$: $3/60$; $x=1$: $17/60$; $x=2$: $40/60$ | B1, B1ft **[4]** | Another correct probability; ft if only one prob correct and $\Sigma p = 1$ |7 Box $A$ contains 5 red paper clips and 1 white paper clip. Box $B$ contains 7 red paper clips and 2 white paper clips. One paper clip is taken at random from box $A$ and transferred to box $B$. One paper clip is then taken at random from box $B$.\\
(i) Find the probability of taking both a white paper clip from box $A$ and a red paper clip from box $B$.\\
(ii) Find the probability that the paper clip taken from box $B$ is red.\\
(iii) Find the probability that the paper clip taken from box $A$ was red, given that the paper clip taken from box $B$ is red.\\
(iv) The random variable $X$ denotes the number of times that a red paper clip is taken. Draw up a table to show the probability distribution of $X$.
\hfill \mbox{\textit{CAIE S1 2007 Q7 [10]}}