| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| 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 and probability distribution construction. Part (a) is a guided calculation, part (b) involves computing three more probabilities using the same method, and part (c) is direct application of the expectation formula. All techniques are standard S1 material with no novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For one yellow: \(YGG + GYG + GGY\); \(\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times 3\) | M1 | \(\frac{a}{9} \times \frac{b}{8} \times \frac{c}{7}\), \(0 < a,b,c\) integers \(\leq 5\), for one arrangement |
| M1 | *Their* three-factor probability \(\times 3\), \(^3C_1\), \(^3C_2\) or \(^3P_1\) (or repeated adding) no additional terms | |
| \(\left[\frac{180}{504} =\right] \frac{5}{14}\) | A1 | AG. Convincingly shown, including identifying possible scenarios, may be on tree diagram WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{^5C_1 \times\, ^4C_2}{^9C_3}\) | M1 | \(\dfrac{^5C_1 \times\, ^4C_2}{^9C_r}\), \(r = 2, 3, 4\) |
| M1 | \(\dfrac{^5C_s \times\, ^4C_t}{^9C_3}\), \(s + t = 3\) | |
| \(\left[\frac{30}{84} =\right] \frac{5}{14}\) | A1 | AG. Convincingly shown, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X\): \(0, 1, 2, 3\); \(P(X)\): \(\frac{24}{504}, \frac{180}{504}, \frac{240}{504}, \frac{60}{504}\) \(\left[= \frac{1}{21}, \frac{5}{14}, \frac{10}{21}, \frac{5}{42}\right]\) \([0.0476, 0.357, 0.476, 0.119]\) | B1 | Table with correct \(X\) values and one correct probability inserted appropriately. Condone any additional \(X\) values if probability stated as 0 |
| B1 | Second identified correct probability, may not be in table | |
| B1 | All probabilities identified and correct. SC if less than 2 correct probabilities or \(X\) value(s) omitted: SC B1 3 or 4 probabilities summing to one |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([E(X) =] \frac{840}{504}, \frac{5}{3}, 1.67\) | B1 | OE Must be evaluated. SC B1 FT correct unsimplified expression from incorrect 3(b) using at least 3 probabilities, \(0 < p < 1\) |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For one yellow: $YGG + GYG + GGY$; $\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times 3$ | M1 | $\frac{a}{9} \times \frac{b}{8} \times \frac{c}{7}$, $0 < a,b,c$ integers $\leq 5$, for one arrangement |
| | M1 | *Their* three-factor probability $\times 3$, $^3C_1$, $^3C_2$ or $^3P_1$ (or repeated adding) no additional terms |
| $\left[\frac{180}{504} =\right] \frac{5}{14}$ | A1 | AG. Convincingly shown, including identifying possible scenarios, may be on tree diagram WWW |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{^5C_1 \times\, ^4C_2}{^9C_3}$ | M1 | $\dfrac{^5C_1 \times\, ^4C_2}{^9C_r}$, $r = 2, 3, 4$ |
| | M1 | $\dfrac{^5C_s \times\, ^4C_t}{^9C_3}$, $s + t = 3$ |
| $\left[\frac{30}{84} =\right] \frac{5}{14}$ | A1 | AG. Convincingly shown, WWW |
**Total: 3 marks**
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X$: $0, 1, 2, 3$; $P(X)$: $\frac{24}{504}, \frac{180}{504}, \frac{240}{504}, \frac{60}{504}$ $\left[= \frac{1}{21}, \frac{5}{14}, \frac{10}{21}, \frac{5}{42}\right]$ $[0.0476, 0.357, 0.476, 0.119]$ | B1 | Table with correct $X$ values and one correct probability inserted appropriately. Condone any additional $X$ values if probability stated as 0 |
| | B1 | Second identified correct probability, may not be in table |
| | B1 | All probabilities identified and correct. **SC** if less than 2 correct probabilities or $X$ value(s) omitted: **SC B1** 3 or 4 probabilities summing to one |
**Total: 3 marks**
---
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[E(X) =] \frac{840}{504}, \frac{5}{3}, 1.67$ | B1 | OE Must be evaluated. **SC B1 FT** correct unsimplified expression from incorrect **3(b)** using at least 3 probabilities, $0 < p < 1$ |
**Total: 1 mark**
---3 A bag contains 5 yellow and 4 green marbles. Three marbles are selected at random from the bag, without replacement.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that exactly one of the marbles is yellow is $\frac { 5 } { 14 }$.\\
The random variable $X$ is the number of yellow marbles selected.
\item Draw up the probability distribution table for $X$.
\item Find $\mathrm { E } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q3 [7]}}