| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Complete probability distribution table |
| Difficulty | Moderate -0.3 This is a straightforward hypergeometric distribution question requiring systematic calculation of probabilities using combinations. Part (i) guides students through one calculation, and part (ii) requires completing the table for all possible values (X=0,1,2,3). While it involves multiple calculations, the method is standard and the numbers are small enough to compute without difficulty. Slightly easier than average due to the scaffolding in part (i) and routine nature of the task. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(2) = {}^6C_3 \times {}^3C_2 / {}^9C_5\) | M1 | Using combinations \({}^aC_b \times {}^cC_d / {}^eC_f\) |
| OR \(\dfrac{{}^6C_3 \times {}^3C_2}{{}^6C_5 + {}^6C_4 \times {}^3C_1 + {}^6C_3 \times {}^3C_2 + {}^6C_2 \times {}^3C_3}\) | M1 | |
| OR \(3/9 \times 2/8 \times 6/7 \times 5/6 \times 4/5 \times {}^5C_2 = 10/21\) | M1 | Mult 5 probs with a \({}^nC_q\); if \({}^5C_2\) replace by 10, oe must be justified |
| \(= 60/126\) AG | A1 | Legit method, as answer given — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Table with \(x\): 0, 1, 2, 3 only seen | B1 | Condone \(x = 4,5\) in table if \(P(x)=0\) or blank and values in table for \(x=0,1,2,3\) |
| \(P(0) = {}^6C_5/{}^9C_5 = 6/126\) | B1 | Any correct prob other than \(P(2)\) |
| \(P(1) = {}^6C_4 \times {}^3C_1/{}^9C_5 = 45/126\) | B1 | Any other correct prob |
| \(P(3) = {}^6C_2 \times {}^3C_3/126 = 15/126\) | B1\(\checkmark\) | \(\Sigma P(x)=1\), \(3 < n(x) < 6\) — Total: 4 |
## Question 3:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(2) = {}^6C_3 \times {}^3C_2 / {}^9C_5$ | M1 | Using combinations ${}^aC_b \times {}^cC_d / {}^eC_f$ |
| **OR** $\dfrac{{}^6C_3 \times {}^3C_2}{{}^6C_5 + {}^6C_4 \times {}^3C_1 + {}^6C_3 \times {}^3C_2 + {}^6C_2 \times {}^3C_3}$ | M1 | |
| **OR** $3/9 \times 2/8 \times 6/7 \times 5/6 \times 4/5 \times {}^5C_2 = 10/21$ | M1 | Mult 5 probs with a ${}^nC_q$; if ${}^5C_2$ replace by 10, oe must be justified |
| $= 60/126$ **AG** | A1 | Legit method, as answer given — **Total: 2** |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Table with $x$: 0, 1, 2, 3 only seen | B1 | Condone $x = 4,5$ in table if $P(x)=0$ or blank and values in table for $x=0,1,2,3$ |
| $P(0) = {}^6C_5/{}^9C_5 = 6/126$ | B1 | Any correct prob other than $P(2)$ |
| $P(1) = {}^6C_4 \times {}^3C_1/{}^9C_5 = 45/126$ | B1 | Any other correct prob |
| $P(3) = {}^6C_2 \times {}^3C_3/126 = 15/126$ | B1$\checkmark$ | $\Sigma P(x)=1$, $3 < n(x) < 6$ — **Total: 4** |
---3 A pet shop has 6 rabbits and 3 hamsters. 5 of these pets are chosen at random. The random variable $X$ represents the number of hamsters chosen.\\
(i) Show that the probability that exactly 2 hamsters are chosen is $\frac { 10 } { 21 }$.\\
(ii) Draw up the probability distribution table for $X$.
\hfill \mbox{\textit{CAIE S1 2014 Q3 [6]}}