| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sampling without replacement |
| Difficulty | Moderate -0.8 This is a straightforward sampling without replacement problem requiring basic probability calculations (combinations or tree diagrams), constructing a simple probability distribution table, and computing an expectation using the standard formula. All techniques are routine for S1 level with no novel insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(1\text{ W}) = \frac{6}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{6}{8}\) | M1 | Summing 2 two-factor probs (condone replacement) not \(\frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2}\) |
| \(= \frac{1}{2}\) AG | A1 [2] | Correct answer, fully justified |
| OR \(\frac{{}^6C_1 \times {}^3C_1}{{}^9C_2}\) | M1 | Using combinations consistent, correct format |
| \(= \frac{1}{2}\) AG | A1 | Correct answer, fully justified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\overline{W}, \overline{W}) = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72}\ (\frac{1}{12})\) | B1 | Distribution table with 0,1,2 only |
| \(P(W, W) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72}\ (\frac{5}{12})\) | B1 | \(P(W,W)\) or \(P(\overline{W},\overline{W})\) correct |
| \(x\): \(0\), \(1\), \(2\); Prob: \(\frac{1}{12}\), \(\frac{1}{2}\), \(\frac{5}{12}\) | B1\(\checkmark\) [3] | \(P(W,W) + P(\overline{W},\overline{W}) = 0.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \frac{16}{12}\ (\frac{4}{3})\ (1.33)\) isw | B1 [1] | Condone \(1(.3)\) if correct working seen, nfww |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(1\text{ W}) = \frac{6}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{6}{8}$ | M1 | Summing 2 two-factor probs (condone replacement) not $\frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2}$ |
| $= \frac{1}{2}$ AG | A1 **[2]** | Correct answer, fully justified |
| OR $\frac{{}^6C_1 \times {}^3C_1}{{}^9C_2}$ | M1 | Using combinations consistent, correct format |
| $= \frac{1}{2}$ AG | A1 | Correct answer, fully justified |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\overline{W}, \overline{W}) = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72}\ (\frac{1}{12})$ | B1 | Distribution table with 0,1,2 only |
| $P(W, W) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72}\ (\frac{5}{12})$ | B1 | $P(W,W)$ or $P(\overline{W},\overline{W})$ correct |
| $x$: $0$, $1$, $2$; Prob: $\frac{1}{12}$, $\frac{1}{2}$, $\frac{5}{12}$ | B1$\checkmark$ **[3]** | $P(W,W) + P(\overline{W},\overline{W}) = 0.5$ |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \frac{16}{12}\ (\frac{4}{3})\ (1.33)$ isw | B1 **[1]** | Condone $1(.3)$ if correct working seen, nfww |
---4 A pet shop has 9 rabbits for sale, 6 of which are white. A random sample of two rabbits is chosen without replacement.\\
(i) Show that the probability that exactly one of the two rabbits in the sample is white is $\frac { 1 } { 2 }$.\\
(ii) Construct the probability distribution table for the number of white rabbits in the sample.\\
(iii) Find the expected value of the number of white rabbits in the sample.
\hfill \mbox{\textit{CAIE S1 2015 Q4 [6]}}