| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sequential trials until success |
| Difficulty | Moderate -0.3 This is a straightforward hypergeometric/sequential sampling problem requiring systematic enumeration of cases. Part (i) is a simple verification (P(RRB) = 2/6 × 1/5 × 4/4), and part (ii) requires calculating P(X=1), P(X=2), P(X=3) using basic conditional probability. The context is clear, calculations are routine, and it's a standard S1 textbook exercise with no conceptual subtlety—slightly easier than average due to small numbers and direct application of sampling without replacement. |
| Spec | 5.02a Discrete probability distributions: general |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=3) = P(RRB) = \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4}\) | (M1) | Probabilities in order \(\frac{2}{p} \times \frac{1}{q} \times \frac{4}{r}\), \(p,q,r \leqslant 6\) and \(p \geqslant q \geqslant r,\ r \geqslant 4\); accept \(\times 1\) as \(\frac{4}{r}\) |
| \(= \frac{1}{15}\) AG | A1) | Needs either P(RRB) OE stated or identified on tree diagram. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=3) = P(RRB) = \frac{^2C_2}{^6C_2} \times \frac{^4C_1}{^4C_1}\) | (M1) | Probabilities stated clearly, \(\times \frac{^4C_1}{^4C_1}\) or \(\times 1\) or \(\times \frac{4}{4}\) included |
| \(= \frac{1}{15}\) AG | A1) | Needs either P(RRB) OE stated or identified on tree diagram. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=3) = P(RRB) = \frac{^2C_1}{^6C_1} \times \frac{^1C_1}{^5C_1} \times \frac{^4C_1}{^4C_1}\) | (M1) | Probabilities in order \(\frac{^2C_1}{^pC_1} \times \frac{^1C_1}{^qC_1} \times \frac{^4C_1}{^rC_1}\), \(p,q,r \leqslant 6\) and \(p \geqslant q \geqslant r,\ r \geqslant 4\); \(\left(\times \frac{^4C_1}{^4C_1}\right.\) or \(\times 1\) or \(\times \frac{4}{4}\) acceptable\()\) |
| \(= 1/15\) AG | A1) | Needs either P(RRB) OE stated or identified on tree diagram. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(1) = P(B) = \frac{4}{6} = \frac{2}{3} = 0.667\) | B1 | Probability distribution table drawn with at least 2 correct \(x\) values and at least 1 probability. All probabilities \(0 \leq p < 1\) |
| \(P(2) = P(RB) = \frac{2}{6} \times \frac{4}{5} = \frac{4}{15} (= 0.267)\) | B1 | \(P(1)\) or \(P(2)\) correct unsimplified, or better, and identified |
| \(P(3) = P(RRB) = \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4} = \frac{1}{15} (= 0.0667)\) | B1 | All probabilities in table, evaluated correctly OE. Additional \(x\) values must have a stated probability of 0 |
| Table: \(x\): 1, 2, 3; \(P\): \(\frac{10}{15}\), \(\frac{4}{15}\), \(\frac{1}{15}\) | 3 total |
## Question 3(i):
**EITHER:**
$P(X=3) = P(RRB) = \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4}$ | **(M1)** | Probabilities in order $\frac{2}{p} \times \frac{1}{q} \times \frac{4}{r}$, $p,q,r \leqslant 6$ and $p \geqslant q \geqslant r,\ r \geqslant 4$; accept $\times 1$ as $\frac{4}{r}$
$= \frac{1}{15}$ AG | **A1)** | Needs either P(RRB) OE stated or identified on tree diagram.
**OR1:**
$P(X=3) = P(RRB) = \frac{^2C_2}{^6C_2} \times \frac{^4C_1}{^4C_1}$ | **(M1)** | Probabilities stated clearly, $\times \frac{^4C_1}{^4C_1}$ or $\times 1$ or $\times \frac{4}{4}$ included
$= \frac{1}{15}$ AG | **A1)** | Needs either P(RRB) OE stated or identified on tree diagram.
**OR2:**
$P(X=3) = P(RRB) = \frac{^2C_1}{^6C_1} \times \frac{^1C_1}{^5C_1} \times \frac{^4C_1}{^4C_1}$ | **(M1)** | Probabilities in order $\frac{^2C_1}{^pC_1} \times \frac{^1C_1}{^qC_1} \times \frac{^4C_1}{^rC_1}$, $p,q,r \leqslant 6$ and $p \geqslant q \geqslant r,\ r \geqslant 4$; $\left(\times \frac{^4C_1}{^4C_1}\right.$ or $\times 1$ or $\times \frac{4}{4}$ acceptable$)$
$= 1/15$ AG | **A1)** | Needs either P(RRB) OE stated or identified on tree diagram.
**Total: 2 marks**
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(1) = P(B) = \frac{4}{6} = \frac{2}{3} = 0.667$ | B1 | Probability distribution table drawn with at least 2 correct $x$ values and at least 1 probability. All probabilities $0 \leq p < 1$ |
| $P(2) = P(RB) = \frac{2}{6} \times \frac{4}{5} = \frac{4}{15} (= 0.267)$ | B1 | $P(1)$ or $P(2)$ correct unsimplified, or better, and identified |
| $P(3) = P(RRB) = \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4} = \frac{1}{15} (= 0.0667)$ | B1 | All probabilities in table, evaluated correctly OE. Additional $x$ values must have a stated probability of 0 |
| Table: $x$: 1, 2, 3; $P$: $\frac{10}{15}$, $\frac{4}{15}$, $\frac{1}{15}$ | **3 total** | |
---3 A box contains 6 identical-sized discs, of which 4 are blue and 2 are red. Discs are taken at random from the box in turn and not replaced. Let $X$ be the number of discs taken, up to and including the first blue one.\\
(i) Show that $\mathrm { P } ( X = 3 ) = \frac { 1 } { 15 }$.\\
(ii) Draw up the probability distribution table for $X$.\\
\hfill \mbox{\textit{CAIE S1 2017 Q3 [5]}}