| Exam Board | OCR |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2018 |
| Session | March |
| Marks | 8 |
| Topic | Hypergeometric Distribution |
| Type | Conditional probability in hypergeometric context |
| Difficulty | Standard +0.8 This question requires understanding of hypergeometric distribution and conditional probability with multiple parts requiring careful combinatorial reasoning. Part (i) is standard hypergeometric, part (ii) requires conditional probability with hypergeometric distributions (non-trivial), and part (iii) involves alternating arrangements which requires insight into valid sequences. The multi-step nature and need to correctly set up conditional probability in a hypergeometric context elevates this above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(^{10}C_1 \times \,^{30}C_7 = [120 \times 2015800]\) | M1\* | 3.1b |
| \(\div \,^{40}C_{10} = [847660528]\) | DepM1 | 1.1 |
| \(= 0.2882\ldots\) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(^5C_3 \times \,^{15}C_7 = [10 \times 6435]\) | M1\* | 3.4 |
| \(\div \,^{20}C_{10} = [184756]\) | depM1 | 1.1 |
| \(= 0.348297\ldots\) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Number of arrangements \(= \frac{10!}{5!5!} = [252]\) | M1 | 1.1a |
| \(2 \div 252 = \frac{1}{126}\) | A1 | 1.1 |
### Part (i)
$^{10}C_1 \times \,^{30}C_7 = [120 \times 2015800]$ | **M1\*** | 3.1b | One correct $^nC_r$ in numerator
$\div \,^{40}C_{10} = [847660528]$ | **DepM1** | 1.1 | Fully correct expression
$= 0.2882\ldots$ | **A1** | 1.1 | Answer, awrt 0.288
### Part (ii)
$^5C_3 \times \,^{15}C_7 = [10 \times 6435]$ | **M1\*** | 3.4 | One correct $^nC_r$ in numerator
$\div \,^{20}C_{10} = [184756]$ | **depM1** | 1.1 | Fully correct expression
$= 0.348297\ldots$ | **A1** | 1.1 | Answer, 0.348 or better | Or $^{10}C_3 \times \,^{10}C_2 \div \,^{20}C_5$
### Part (iii)
Number of arrangements $= \frac{10!}{5!5!} = [252]$ | **M1** | 1.1a
$2 \div 252 = \frac{1}{126}$ | **A1** | 1.1 | Or exact equivalent, or 0.00794 or better
---3 A pack of 40 cards consists of 10 cards in each of four colours: red, yellow, blue and green. The pack is dealt at random into four "hands", each of 10 cards. The hands are labelled North, South, East and West.\\
(i) Find the probability that West has exactly 3 red cards.\\
(ii) Find the probability that West has exactly 3 red cards, given that East and West have between them a total of exactly 5 red cards.\\
(iii) South has 5 red cards and 5 blue cards. These cards are placed in a row in a random order. Find the probability that the colour of each card is different from the colour of the preceding card.
\hfill \mbox{\textit{OCR FS1 AS 2018 Q3 [8]}}