OCR MEI S1 2012 June — Question 1 6 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypergeometric Distribution
TypeAt least one of each type
DifficultyModerate -0.8 This is a straightforward application of the hypergeometric distribution formula with small numbers and only two parts. Part (i) requires a single probability calculation, while part (ii) needs the complement rule but still involves routine computation. The context is clear and the calculations are mechanical with no conceptual challenges beyond recognizing the sampling-without-replacement scenario.
Spec5.01a Permutations and combinations: evaluate probabilities

1 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.
  1. Find the probability that all 3 of these bulbs will produce blue flowers.
  2. Find the probability that Karen will have at least one flower of each colour from her 3 bulbs.

(First page is not shown in detail but appears to be continuation of probability/statistics content)
AnswerMarks
M1For sum of both or for 3×
M1NB M2 also for
A1CAO
[3]
M1For either \(^nC_3\) or \(^nC_3\)
M1For product of both
A1CAO
[3]
M1For \(^nC_6\) seen in part (ii)
M1For their 840/3003 or their 840/\(^nC_6\)
A1FT their 840
[3]Allow full marks for unsimplified fractional answers
NOTE RE OVER-SPECIFICATION OF ANSWERS
If answers are grossly over-specified, deduct the final answer mark in every case. Probabilities should also be rounded to a sensible degree of accuracy. In general final non-probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. In general accept answers which are correct to 3 significant figures when given to 4 or 5 significant figures.
If answer given as a fraction and as an over-specified decimal – ignore decimal and mark fraction.
(First page is not shown in detail but appears to be continuation of probability/statistics content)

| | M1 | For sum of both or for 3× |
| | M1 | NB M2 also for |
| | A1 | CAO |
| | [3] | |
| | | |
| | M1 | For either $^nC_3$ or $^nC_3$ |
| | M1 | For product of both |
| | A1 | CAO |
| | [3] | |
| | | |
| | M1 | For $^nC_6$ seen in part (ii) |
| | M1 | For their 840/3003 or their 840/$^nC_6$ |
| | A1 | FT their 840 |
| | [3] | Allow full marks for unsimplified fractional answers |

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# NOTE RE OVER-SPECIFICATION OF ANSWERS

If answers are grossly over-specified, deduct the final answer mark in every case. Probabilities should also be rounded to a sensible degree of accuracy. In general final non-probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. In general accept answers which are correct to 3 significant figures when given to 4 or 5 significant figures.

If answer given as a fraction and as an over-specified decimal – ignore decimal and mark fraction.
1 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 2012 Q1 [6]}}