| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | At least one of each type |
| Difficulty | Moderate -0.3 This is a straightforward hypergeometric distribution problem requiring direct application of the formula or systematic counting. Part (i) is a single calculation (30C3)/(50C3), and part (ii) requires finding P(at least one of each) = 1 - P(all blue) - P(all red), involving three simple calculations. The context is clear, the method is standard, and no conceptual insight beyond recognizing the sampling-without-replacement scenario is needed. |
| Spec | 2.03a Mutually exclusive and independent events |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{All blue}) = \frac{30}{50} \times \frac{29}{49} \times \frac{28}{48} = 0.2071\) | M1 | For \(\frac{30}{50}\) as part of a triple product; \((30/50)^3 = 0.216\) scores M1M0A0; \(\frac{k}{50} \times \frac{(k-1)}{49} \times \frac{(k-2)}{48}\) for values of \(k\) other than 30 scores M1M0A0; Zero for binomial unless simplifies to \((3/5)^3\) |
| Product of other two fractions \(\frac{29}{49} \times \frac{28}{48}\) | M1 | Correct working but then multiplied or divided by some factor scores M1M0A0 |
| \(\binom{30}{3}/\binom{50}{3} = 4060/19600 = 29/140 = 0.2071\) | A1 | CAO; Accept 0.21 with working and 0.207 without working; Allow unsimplified fraction as final answer \(24360/117600\) oe |
| SC2 for \(P(\text{All red}) = 0.0582\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{All red}) = \frac{20}{50} \times \frac{19}{49} \times \frac{18}{48} = 0.0582\) or \(\binom{20}{3}/\binom{50}{3} = 0.0582\) | M1 | For \(P(\text{All red})\); SC2 for \(1-(30/50)^3-(20/50)^3 = 1-0.216-0.064=0.72\), providing consistent with (i); If not consistent with (i) M0M0A0 |
| \(P(\text{At least one of each colour}) = 1-(0.2071+0.0582) = 0.7347\) or \(1-\left(\frac{29}{140}+\frac{57}{980}\right)=1-\frac{260}{980}=1-\frac{13}{49}=\frac{36}{49}\) | M1 | For \(1-(0.2071+0.0582)\) |
| \(0.7347\) | A1 | CAO; Allow 0.73 with working; Allow unsimplified fraction as final answer \(86400/117600\) oe |
| OR: \(P(2b,1r)+P(1b,2r)\) | (M1) | For either \(\frac{30}{50}\times\frac{29}{49}\times\frac{20}{48}\) or \(\frac{20}{50}\times\frac{19}{49}\times\frac{30}{48}\); Allow M1 for \(3\times(30/50)^2\times(20/50)\) or \(3\times(30/50)\times(20/50)^2\) and second M1 for sum of both if \(=0.72\); If not consistent with (i) M0M0A0 |
## Question 7:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{All blue}) = \frac{30}{50} \times \frac{29}{49} \times \frac{28}{48} = 0.2071$ | M1 | For $\frac{30}{50}$ as part of a triple product; $(30/50)^3 = 0.216$ scores M1M0A0; $\frac{k}{50} \times \frac{(k-1)}{49} \times \frac{(k-2)}{48}$ for values of $k$ other than 30 scores M1M0A0; Zero for binomial unless simplifies to $(3/5)^3$ |
| Product of other two fractions $\frac{29}{49} \times \frac{28}{48}$ | M1 | Correct working but then multiplied or divided by some factor scores M1M0A0 |
| $\binom{30}{3}/\binom{50}{3} = 4060/19600 = 29/140 = 0.2071$ | A1 | CAO; Accept 0.21 with working and 0.207 without working; Allow unsimplified fraction as final answer $24360/117600$ oe |
| SC2 for $P(\text{All red}) = 0.0582$ | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{All red}) = \frac{20}{50} \times \frac{19}{49} \times \frac{18}{48} = 0.0582$ or $\binom{20}{3}/\binom{50}{3} = 0.0582$ | M1 | For $P(\text{All red})$; SC2 for $1-(30/50)^3-(20/50)^3 = 1-0.216-0.064=0.72$, providing consistent with (i); If not consistent with (i) M0M0A0 |
| $P(\text{At least one of each colour}) = 1-(0.2071+0.0582) = 0.7347$ or $1-\left(\frac{29}{140}+\frac{57}{980}\right)=1-\frac{260}{980}=1-\frac{13}{49}=\frac{36}{49}$ | M1 | For $1-(0.2071+0.0582)$ |
| $0.7347$ | A1 | CAO; Allow 0.73 with working; Allow unsimplified fraction as final answer $86400/117600$ oe |
| OR: $P(2b,1r)+P(1b,2r)$ | (M1) | For either $\frac{30}{50}\times\frac{29}{49}\times\frac{20}{48}$ or $\frac{20}{50}\times\frac{19}{49}\times\frac{30}{48}$; Allow M1 for $3\times(30/50)^2\times(20/50)$ or $3\times(30/50)\times(20/50)^2$ and second M1 for sum of both if $=0.72$; If not consistent with (i) M0M0A0 |
---7 At a garden centre there is a box containing 50 hyacinth bulbs. Of these, 30 will produce a blue flower and the remaining 20 will produce a red flower. Unfortunately they have become mixed together so that it is not known which of the bulbs will produce a blue flower and which will produce a red flower.
Karen buys 3 of these bulbs.\\
(i) Find the probability that all 3 of these bulbs will produce blue flowers.\\
(ii) Find the probability that Karen will have at least one flower of each colour from her 3 bulbs.
\hfill \mbox{\textit{OCR MEI S1 Q7 [6]}}