| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Sequential selection without replacement |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram question with a simple twist (adding a yellow ball). Part (i) requires basic probability calculations with changing totals. Parts (ii) and (iii) involve standard probability rules (multiplication and addition for part ii, conditional probability for part iii). The question is easier than average because it's a routine application of well-practiced techniques with no conceptual surprises, though the yellow ball addition prevents it from being completely trivial. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram with First Ball: R (3/8), B (5/8); Second Ball given R: R (2/8), B (5/8), Y (1/8); given B: R (3/8), B (4/8), Y (1/8) | B1 | Fully correct labelled tree and correct probabilities for 'First Ball' |
| B1 | Correct probabilities (with corresponding labels) for 'Second Ball' |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(RR) + P(BB) = \frac{3}{8} \times \frac{2}{8} + \frac{5}{8} \times \frac{4}{8} = \frac{3}{32} + \frac{5}{16}\) | M1 | Correct unsimplified expression from their tree diagram, \(\Sigma p = 1\) on each branch |
| \(= 13/32\ (0.406)\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(RB) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}\) | M1 | \(P(\text{1st ball red}) \times P(\text{2nd ball blue})\) from their tree diagram seen unsimplified as numerator or denominator of a fraction. Allow \(\Sigma p \neq 1\) on each branch |
| \(P(B) = \frac{3}{8} \times \frac{5}{8} + \frac{5}{8} \times \frac{4}{8} = \frac{35}{64}\) | M1 | Correct unsimplified expression for \(P(B)\) from their tree diagram seen as denominator of a fraction. Allow \(\Sigma p \neq 1\) on each branch |
| \(P(R | B) = P(RB)/P(B) = (15/64) \div (35/64) = 3/7\ (0.429)\) | A1 |
## Question 3(i):
Tree diagram with First Ball: R (3/8), B (5/8); Second Ball given R: R (2/8), B (5/8), Y (1/8); given B: R (3/8), B (4/8), Y (1/8) | B1 | Fully correct labelled tree and correct probabilities for 'First Ball'
| B1 | Correct probabilities (with corresponding labels) for 'Second Ball'
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## Question 3(ii):
$P(RR) + P(BB) = \frac{3}{8} \times \frac{2}{8} + \frac{5}{8} \times \frac{4}{8} = \frac{3}{32} + \frac{5}{16}$ | M1 | Correct unsimplified expression from their tree diagram, $\Sigma p = 1$ on each branch
$= 13/32\ (0.406)$ | A1 | Correct answer
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## Question 3(iii):
$P(RB) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}$ | M1 | $P(\text{1st ball red}) \times P(\text{2nd ball blue})$ from their tree diagram seen unsimplified as numerator or denominator of a fraction. Allow $\Sigma p \neq 1$ on each branch
$P(B) = \frac{3}{8} \times \frac{5}{8} + \frac{5}{8} \times \frac{4}{8} = \frac{35}{64}$ | M1 | Correct unsimplified expression for $P(B)$ from their tree diagram seen as denominator of a fraction. Allow $\Sigma p \neq 1$ on each branch
$P(R|B) = P(RB)/P(B) = (15/64) \div (35/64) = 3/7\ (0.429)$ | A1 | Correct answer
---3 A box contains 3 red balls and 5 blue balls. One ball is taken at random from the box and not replaced. A yellow ball is then put into the box. A second ball is now taken at random from the box.\\
(i) Complete the tree diagram to show all the outcomes and the probability for each branch.
First ball\\
Second ball\\
\includegraphics[max width=\textwidth, alt={}, center]{7dc85f33-2647-4f73-8093-524b70f99767-04_655_392_688_474}\\
\includegraphics[max width=\textwidth, alt={}, center]{7dc85f33-2647-4f73-8093-524b70f99767-04_785_387_703_1110}\\
(ii) Find the probability that the two balls taken are the same colour.\\
(iii) Find the probability that the first ball taken is red, given that the second ball taken is blue.\\
\hfill \mbox{\textit{CAIE S1 2018 Q3 [7]}}