| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Verify or show equation |
| Difficulty | Moderate -0.8 This is a straightforward hypergeometric distribution question requiring basic probability calculations without replacement. Part (a) is a simple 'show that' verification using combinations, part (b) requires calculating three similar probabilities, and part (c) uses the variance formula with E(X) given. All steps are routine applications of standard formulas with no problem-solving insight needed. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{1 red}) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times 3\) | M1 | \(\frac{a}{8} \times \frac{b}{7} \times \frac{c}{6} \times k\) or \(\frac{5}{d} \times \frac{3}{e} \times \frac{2}{f} \times 3\), \(1 \leq a,b,c \leq 5\), \(d,e,f \leq 8\), \(a,b,c,d,e,f,k\) all integers, \(1 < k \leq 3\) |
| \(\frac{15}{56}\) | A1 | AG, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{{}^5C_1 \times {}^3C_2}{{}^8C_3}\) | M1 | \(\frac{{}^aC_1 \times {}^bC_2}{{}^8C_3}\) or \(\frac{{}^5C_d \times {}^3C_e}{{}^8C_3}\); \(a+b=8\), \(d+e=3\) |
| \(\frac{15}{56}\) | A1 | AG, WWW, \(\frac{15}{56}\) must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | 0 | 1 |
| Prob. | \(\frac{1}{56}\) | \(\frac{15}{56}\) |
| B1 | Probability distribution table with correct outcomes, at least one probability less than 1, allow extra outcome values if probability of zero stated | |
| 2 of \(P(0)\), \(P(2)\) and \(P(3)\) correct | B1 | |
| \(4^{\text{th}}\) probability correct or FT sum of 3 or more probabilities \(= 1\), with \(P(1)\) correct | B1 FT |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(X) = \frac{0^2 \times 1 + 1^2 \times 15 + 2^2 \times 30 + 3^2 \times 10}{56} - \left(\frac{15}{8}\right)^2\) | M1 | Substitute their attempts at scores in correct variance formula, must have \(- \text{mean}^2\) (FT if mean calculated), condone probabilities not summing to 1 |
| Answer | Marks |
|---|---|
| \(\frac{225}{448}\), \(0.502\) | A1 |
## Question 2(a):
$P(\text{1 red}) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times 3$ | M1 | $\frac{a}{8} \times \frac{b}{7} \times \frac{c}{6} \times k$ or $\frac{5}{d} \times \frac{3}{e} \times \frac{2}{f} \times 3$, $1 \leq a,b,c \leq 5$, $d,e,f \leq 8$, $a,b,c,d,e,f,k$ all integers, $1 < k \leq 3$
$\frac{15}{56}$ | A1 | AG, WWW
**Alternative method:**
$\frac{{}^5C_1 \times {}^3C_2}{{}^8C_3}$ | M1 | $\frac{{}^aC_1 \times {}^bC_2}{{}^8C_3}$ or $\frac{{}^5C_d \times {}^3C_e}{{}^8C_3}$; $a+b=8$, $d+e=3$
$\frac{15}{56}$ | A1 | AG, WWW, $\frac{15}{56}$ must be seen
---
## Question 2(b):
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Prob. | $\frac{1}{56}$ | $\frac{15}{56}$ | $\frac{30}{56}=\frac{15}{28}$ | $\frac{10}{56}=\frac{5}{28}$ |
| B1 | Probability distribution table with correct outcomes, at least one probability less than 1, allow extra outcome values if probability of zero stated
2 of $P(0)$, $P(2)$ and $P(3)$ correct | B1 |
$4^{\text{th}}$ probability correct or FT sum of 3 or more probabilities $= 1$, with $P(1)$ correct | B1 FT |
---
## Question 2(c):
$\text{Var}(X) = \frac{0^2 \times 1 + 1^2 \times 15 + 2^2 \times 30 + 3^2 \times 10}{56} - \left(\frac{15}{8}\right)^2$ | M1 | Substitute their attempts at scores in correct variance formula, must have $- \text{mean}^2$ (FT if mean calculated), condone probabilities not summing to 1
$= \frac{15}{56} + \frac{120}{56} + \frac{90}{56} - \left(\frac{15}{8}\right)^2$
$\frac{225}{448}$, $0.502$ | A1 |
---2 A bag contains 5 red balls and 3 blue balls. Sadie takes 3 balls at random from the bag, without replacement. The random variable $X$ represents the number of red balls that she takes.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that Sadie takes exactly 1 red ball is $\frac { 15 } { 56 }$.
\item Draw up the probability distribution table for $X$.
\item Given that $\mathrm { E } ( X ) = \frac { 15 } { 8 }$, find $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q2 [7]}}