| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| 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 combinatorics (choosing 2 from 10) and expectation calculation. Part (i) is guided ('show that'), part (ii) involves computing three simple probabilities, and part (iii) is routine expectation from a discrete distribution. All steps are standard textbook exercises with no novel insight required, making it easier than average. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(N, \overline{N}) = \frac{3}{10} \times \frac{7}{9}\) | M1 | For multiplying 2 relevant possibilities |
| Multiply by 2 = \(\frac{7}{15}\) AG | A1 (2) | For obtaining given answer legitimately |
| OR Total ways \({}_{10}C_2 = 45\); Total 1 of each: \({}_{7}C_1 \times {}_{3}C_1 = 21\) | M1 | For both totals |
| \(\text{Prob} = 21/45 = 7/15\) AG | A1 (2) | For obtaining correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(N, N) - \frac{3}{10} \times \frac{2}{9} \left(= \frac{1}{15}\right)\) | M1 | For 2 correct numbers multiplied together, can be implied |
| \(P(\overline{N}, \overline{N}) = \frac{7}{10} \times \frac{6}{9} \left(= \frac{7}{15}\right)\) | M1 | For 2 correct numbers multiplied together or subtracting from 1 |
| \(\dfrac{x \quad\ | \quad 0 \quad 1 \quad 2}{P(X=x)\ | \; \frac{7}{15} \; \frac{7}{15} \; \frac{1}{15}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = \frac{3}{5}\) | B1ft (1) | For correct answer or equivalent. Only ft if \(\sum p = 1\) |
## Question 2:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(N, \overline{N}) = \frac{3}{10} \times \frac{7}{9}$ | M1 | For multiplying 2 relevant possibilities |
| Multiply by 2 = $\frac{7}{15}$ **AG** | A1 (2) | For obtaining given answer legitimately |
| **OR** Total ways ${}_{10}C_2 = 45$; Total 1 of each: ${}_{7}C_1 \times {}_{3}C_1 = 21$ | M1 | For both totals |
| $\text{Prob} = 21/45 = 7/15$ **AG** | A1 (2) | For obtaining correct answer |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(N, N) - \frac{3}{10} \times \frac{2}{9} \left(= \frac{1}{15}\right)$ | M1 | For 2 correct numbers multiplied together, can be implied |
| $P(\overline{N}, \overline{N}) = \frac{7}{10} \times \frac{6}{9} \left(= \frac{7}{15}\right)$ | M1 | For 2 correct numbers multiplied together or subtracting from 1 |
| $\dfrac{x \quad\|\quad 0 \quad 1 \quad 2}{P(X=x)\| \; \frac{7}{15} \; \frac{7}{15} \; \frac{1}{15}}$ | B1 (3) | All correct. Table correct and no working gets 3/3 |
**(iii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = \frac{3}{5}$ | B1ft (1) | For correct answer or equivalent. Only ft if $\sum p = 1$ |
---2 A box contains 10 pens of which 3 are new. A random sample of two pens is taken.\\
(i) Show that the probability of getting exactly one new pen in the sample is $\frac { 7 } { 15 }$.\\
(ii) Construct a probability distribution table for the number of new pens in the sample.\\
(iii) Calculate the expected number of new pens in the sample.
\hfill \mbox{\textit{CAIE S1 2003 Q2 [6]}}