| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability of specific committee composition |
| Difficulty | Moderate -0.3 This is a straightforward combinations question requiring calculation of C(11,3) and then finding P(at least 2 terriers) using basic probability rules. The multi-part structure and need to consider complementary cases (2 or 3 terriers) adds slight complexity, but the techniques are standard S1 material with no conceptual challenges. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(\binom{11}{3} = 165\) | M1 | Seen |
| \(= 165\) | A1 [2] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{\binom{5}{2}\times\binom{6}{1} + \binom{5}{3}\times\binom{6}{0}}{\binom{11}{3}} = \frac{60}{165} + \frac{10}{165} = \frac{70}{165} = \frac{14}{33} = 0.424\] | M1 | For intention to add correct two fractional terms |
| Numerator of first term | M1 | For numerator of first term |
| Numerator of second term | M1 | For numerator of second term. Do not penalise omission of \(\binom{6}{0}\) |
| Correct denominator | M1 | For correct denominator |
| \(\frac{14}{33}\) | A1 [5] | cao |
## Question 3:
### Part (i):
$\binom{11}{3} = 165$ | M1 | Seen
$= 165$ | A1 [2] | cao
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### Part (ii):
$$\frac{\binom{5}{2}\times\binom{6}{1} + \binom{5}{3}\times\binom{6}{0}}{\binom{11}{3}} = \frac{60}{165} + \frac{10}{165} = \frac{70}{165} = \frac{14}{33} = 0.424$$ | M1 | For intention to add correct two fractional terms | **Or** For attempt at correct two terms
Numerator of first term | M1 | For numerator of first term | For product of 3 correct fractions $= \frac{4}{33}$
Numerator of second term | M1 | For numerator of second term. Do not penalise omission of $\binom{6}{0}$ | For whole expression i.e. $3\times\frac{5}{11}\times\frac{4}{10}\times\frac{6}{9}\left(=\frac{4}{11}\right)(= 3\times 0.1212...)$
Correct denominator | M1 | For correct denominator | For attempt at $\frac{5}{11}\times\frac{4}{10}\times\frac{3}{9}\left(=\frac{2}{33}\right)$
$\frac{14}{33}$ | A1 [5] | cao | cao. Use of binomial can get max first M1
**Alternative:** $1 - P(1 \text{ or } 0) = 1 - 3\times\frac{5}{11}\times\frac{6}{10}\times\frac{5}{9} - \frac{6}{11}\times\frac{5}{10}\times\frac{4}{9}$
$= 1 - \frac{5}{11} - \frac{4}{33} = \frac{14}{33}$
M1 for $1 - P(1 \text{ or } 0)$, M1 for first product, M1 for $\times 3$, M1 for second product, A13 At a dog show, three out of eleven dogs are to be selected for a national competition.\\
(i) Find the number of possible selections.\\
(ii) Five of the eleven dogs are terriers. Assuming that the dogs are selected at random, find the probability that at least two of the three dogs selected for the national competition are terriers.
\hfill \mbox{\textit{OCR MEI S1 Q3 [7]}}