| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Repeated trials with selection |
| Difficulty | Standard +0.3 This is a standard S1 combinations question with straightforward applications of selection formulas and binomial probability. Parts (i)-(iv) require basic combinations and conditional probability, while (v)-(vi) apply binomial distribution - all routine techniques for this level with no novel insight required, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{all jam}) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{1}{22} = 0.04545\) | M1 for \(5 \times 4 \times 3\) or \(\binom{5}{3}\) in numerator; M1 for \(12 \times 11 \times 10\) or \(\binom{12}{3}\) in denominator | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{22} + \frac{1}{55} + \frac{1}{220} = \frac{3}{44} = 0.06818\) | M1 sum of 3 reasonable triples or combinations; M1 triples or combinations correct | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{all different}) = 6 \times \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{3}{11} = 0.2727\) | M1 for 5,4,3; M1 for \(6 \times\) three fractions or \(\binom{12}{3}\) denom | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{all jam} \mid \text{all same}) = \dfrac{\frac{1}{22}}{\frac{3}{44}} = \frac{2}{3}\) | M1 their (i) in numerator; M1 their (ii) in denominator | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{all jam exactly twice}) = \binom{5}{2} \times \left(\frac{1}{22}\right)^2 \times \left(\frac{21}{22}\right)^3 = 0.01797\) | M1 for \(\binom{5}{2} \times \ldots\); M1 for their \(p^2 q^3\) | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{all jam at least once}) = 1 - \left(\frac{21}{22}\right)^5 = 0.2075\) | M1 for their \(q^5\); M1 for \(1 - 5^{\text{th}}\) power | A1 CAO |
## Question 2:
### Part (i):
$P(\text{all jam}) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{1}{22} = 0.04545$ | M1 for $5 \times 4 \times 3$ or $\binom{5}{3}$ in numerator; M1 for $12 \times 11 \times 10$ or $\binom{12}{3}$ in denominator | A1 CAO
### Part (ii):
$P(\text{all same}) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} + \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} + \frac{3}{12} \times \frac{2}{11} \times \frac{1}{10}$
$= \frac{1}{22} + \frac{1}{55} + \frac{1}{220} = \frac{3}{44} = 0.06818$ | M1 sum of 3 reasonable triples or combinations; M1 triples or combinations correct | A1 CAO
### Part (iii):
$P(\text{all different}) = 6 \times \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{3}{11} = 0.2727$ | M1 for 5,4,3; M1 for $6 \times$ three fractions or $\binom{12}{3}$ denom | A1 CAO
### Part (iv):
$P(\text{all jam} \mid \text{all same}) = \dfrac{\frac{1}{22}}{\frac{3}{44}} = \frac{2}{3}$ | M1 their (i) in numerator; M1 their (ii) in denominator | A1 CAO
### Part (v):
$P(\text{all jam exactly twice}) = \binom{5}{2} \times \left(\frac{1}{22}\right)^2 \times \left(\frac{21}{22}\right)^3 = 0.01797$ | M1 for $\binom{5}{2} \times \ldots$; M1 for their $p^2 q^3$ | A1 CAO
### Part (vi):
$P(\text{all jam at least once}) = 1 - \left(\frac{21}{22}\right)^5 = 0.2075$ | M1 for their $q^5$; M1 for $1 - 5^{\text{th}}$ power | A1 CAO
---2 Jane buys 5 jam doughnuts, 4 cream doughnuts and 3 plain doughnuts.\\
On arrival home, each of her three children eats one of the twelve doughnuts. The different kinds of doughnut are indistinguishable by sight and so selection of doughnuts is random.
Calculate the probabilities of the following events.\\
(i) All 3 doughnuts eaten contain jam.\\
(ii) All 3 doughnuts are of the same kind.\\
(iii) The 3 doughnuts are all of a different kind.\\
(iv) The 3 doughnuts contain jam, given that they are all of the same kind.
On 5 successive Saturdays, Jane buys the same combination of 12 doughnuts and her three children eat one each. Find the probability that all 3 doughnuts eaten contain jam on\\
(v) exactly 2 Saturdays out of the 5 ,\\
(vi) at least 1 Saturday out of the 5 .
\hfill \mbox{\textit{OCR MEI S1 Q2 [18]}}