| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Repeated trials with selection |
| Difficulty | Standard +0.3 This is a straightforward S1 combinations question with standard probability calculations (selecting without replacement) followed by routine binomial probability applications. Part (iv) requires conditional probability but is clearly signposted, and parts (v)-(vi) are textbook binomial distribution exercises. The multi-part structure and variety of techniques make it slightly above average, but all components are standard exam fare requiring no novel insight. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks |
|---|---|
| \(P(\text{all 3 jam}) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}\) | M1 A1 A1 |
| Answer | Marks |
|---|---|
| \(P(\text{all same}) = \frac{1}{22} + \frac{4}{12}\times\frac{3}{11}\times\frac{2}{10} + \frac{3}{12}\times\frac{2}{11}\times\frac{1}{10}\) | M1 |
| \(= \frac{1}{22} + \frac{24}{1320} + \frac{6}{1320} = \frac{60+24+6}{1320} = \frac{90}{1320} = \frac{3}{44}\) | A1 A1 |
| Answer | Marks |
|---|---|
| \(P(\text{all different}) = \frac{5 \times 4 \times 3 \times 3!}{12 \times 11 \times 10} = \frac{360}{1320} = \frac{3}{11}\) | M1 A1 A1 |
| Answer | Marks |
|---|---|
| \(P(\text{jam} \mid \text{all same}) = \frac{P(\text{all jam})}{P(\text{all same})} = \frac{\frac{1}{22}}{\frac{3}{44}} = \frac{1}{22} \times \frac{44}{3} = \frac{2}{3}\) | M1 A1 A1 |
| Answer | Marks |
|---|---|
| \(P(X=2) = \binom{5}{2}\left(\frac{1}{22}\right)^2\left(\frac{21}{22}\right)^3\) | M1 A1 |
| \(= 10 \times \frac{1}{484} \times \frac{9261}{10648} = \frac{92610}{5153632} \approx 0.01796\) | A1 |
| Answer | Marks |
|---|---|
| \(P(X \geq 1) = 1 - P(X=0) = 1 - \left(\frac{21}{22}\right)^5\) | M1 A1 |
| \(= 1 - \frac{4084101}{5153632} \approx 1 - 0.7925 = 0.2075\) | A1 |
# Question 8:
**(i)**
$P(\text{all 3 jam}) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}$ | M1 A1 A1 |
**(ii)**
$P(\text{all same}) = \frac{1}{22} + \frac{4}{12}\times\frac{3}{11}\times\frac{2}{10} + \frac{3}{12}\times\frac{2}{11}\times\frac{1}{10}$ | M1 |
$= \frac{1}{22} + \frac{24}{1320} + \frac{6}{1320} = \frac{60+24+6}{1320} = \frac{90}{1320} = \frac{3}{44}$ | A1 A1 |
**(iii)**
$P(\text{all different}) = \frac{5 \times 4 \times 3 \times 3!}{12 \times 11 \times 10} = \frac{360}{1320} = \frac{3}{11}$ | M1 A1 A1 |
**(iv)**
$P(\text{jam} \mid \text{all same}) = \frac{P(\text{all jam})}{P(\text{all same})} = \frac{\frac{1}{22}}{\frac{3}{44}} = \frac{1}{22} \times \frac{44}{3} = \frac{2}{3}$ | M1 A1 A1 |
**(v)**
Let $p = \frac{1}{22}$, $X \sim B(5, \frac{1}{22})$ — wait, $p = \frac{1}{22}$
$P(X=2) = \binom{5}{2}\left(\frac{1}{22}\right)^2\left(\frac{21}{22}\right)^3$ | M1 A1 |
$= 10 \times \frac{1}{484} \times \frac{9261}{10648} = \frac{92610}{5153632} \approx 0.01796$ | A1 |
**(vi)**
$P(X \geq 1) = 1 - P(X=0) = 1 - \left(\frac{21}{22}\right)^5$ | M1 A1 |
$= 1 - \frac{4084101}{5153632} \approx 1 - 0.7925 = 0.2075$ | A1 |8 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 2006 Q8 [18]}}