OCR Further Statistics 2024 June — Question 6 11 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2024
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPermutations & Arrangements
TypeSpecific items together
DifficultyStandard +0.3 This is a multi-part permutations question with identical objects that requires systematic counting. Parts (a)-(c) are routine applications of combinations and arrangements formulas. Part (d) requires more careful case analysis (choosing which 3 yellows form a block and positioning constraints), and part (e) tests conceptual understanding. While thorough, the techniques are standard for Further Maths statistics, making this slightly easier than average overall.
Spec5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems
6 A bag contains 6 identical blue counters and 5 identical yellow counters.
  1. Three counters are selected at random, without replacement. Find the probability that at least two of the counters are blue. All 11 counters are now arranged in a row in a random order.
  2. Find the probability that all the yellow counters are next to each other.
  3. Find the probability that no yellow counter is next to another yellow counter.
  4. Find the probability that the counters are arranged in such a way that both of the following conditions hold.
Question 6:
AnswerMarks Guidance
6(a) [(6C  5) + 6C ]  11C [= (15×5 + 20)  165]
2 3 3
5 C + 5 C  6
or 61  51  59  3 + 61  51  49 , or 1 − 3 2
1 0 1 0 1 1C
3
19
= or 0.5757...
AnswerMarks
33M1
A1
AnswerMarks
[2]2.1
1.15
One term omitted from quotient (e.g. = 0.455 or
11
73 7 23 93
= 0.212 or = 0.636 or = 0.879), or one
3 11
31
number from product, (e.g. = 0.273): M1A0
1
Exact or 0.576 or better
Other methods: Correct answer B2, else 0
AnswerMarks
(b)7  11C [= 7  462]
5
16
= or 0.01515...
AnswerMarks
6M1
A1
AnswerMarks
[2]1.1
2.2aOr 51  41  39  28  17  7 (must have 6 or 7)
1 0
6  11C (= 17 = 0.0130) or 7  11P : M1A0
5 7 5
Exact or 0.015 or better, www
Other methods: Correct answer B2, else 0
AnswerMarks
(c)7C  11C [= 21  462]
5 5
12
= or 0.04545...
AnswerMarks
2M1
A1
A1
AnswerMarks
[3]3.1b
1.1
AnswerMarks
2.2aEither correct term seen, e.g. (5!6!)/11!
Fully correct expression
Exact or 0.045 or better, www
Other methods: Correct answer B3, else 0
AnswerMarks
(d)7  6C  11C
2 5
5
= or 0.227272...
AnswerMarks
22M1
A1
A1
AnswerMarks
[3]3.1b
1.1
AnswerMarks
2.2a7P , 7C ,6C , 6P or 7C in numerator, e.g. 765 or
2 2 2 2 3
6×5×4 (e.g. 1 , 1 , 5 , 5 , 5 ) or 56
11 22 154 11 66 6
or diagram showing 6 blues and 7 gaps: M1
Numerator correct
Exact or 0.227 or better
Other methods: Correct answer B3, else 0
AnswerMarks
(e)No because numerator and denominator would both be
multiplied by 5! [factor needn’t be present or correct]
AnswerMarks Guidance
or No as we are concerned only about the colour, OEB1
[1]2.4 No with relevant reason that is not wrong, e.g. “it
makes no difference that they now have labels”. More
than 1 reason: OK unless one is wrong
AnswerMarks Guidance
Not changed as you still treat them as yellowB1
Different ways of arranging counters does not affect probabilityB1
Not changed as being yellow is independent of numbersB1
No as do not require a specific order of yellows or bluesB1
No as still 5 yellow counters to choose fromB1
No as it is only the colour that mattersB1
Both top and bottom are  same number so unchangedB1
No as we use permutations not combinationsB1
No because still the same number of possibilitiesB1
“No”, no reasonB0
QuestionAnswer Marks 
Question 6:
6 | (a) | [(6C  5) + 6C ]  11C [= (15×5 + 20)  165]
2 3 3
5 C + 5 C  6
or 61  51  59  3 + 61  51  49 , or 1 − 3 2
1 0 1 0 1 1C
3
19
= or 0.5757...
33 | M1
A1
[2] | 2.1
1.1 | 5
One term omitted from quotient (e.g. = 0.455 or
11
73 7 23 93
= 0.212 or = 0.636 or = 0.879), or one
3 11
31
number from product, (e.g. = 0.273): M1A0
1
Exact or 0.576 or better
Other methods: Correct answer B2, else 0
(b) | 7  11C [= 7  462]
5
16
= or 0.01515...
6 | M1
A1
[2] | 1.1
2.2a | Or 51  41  39  28  17  7 (must have 6 or 7)
1 0
6  11C (= 17 = 0.0130) or 7  11P : M1A0
5 7 5
Exact or 0.015 or better, www
Other methods: Correct answer B2, else 0
(c) | 7C  11C [= 21  462]
5 5
12
= or 0.04545...
2 | M1
A1
A1
[3] | 3.1b
1.1
2.2a | Either correct term seen, e.g. (5!6!)/11!
Fully correct expression
Exact or 0.045 or better, www
Other methods: Correct answer B3, else 0
(d) | 7  6C  11C
2 5
5
= or 0.227272...
22 | M1
A1
A1
[3] | 3.1b
1.1
2.2a | 7P , 7C ,6C , 6P or 7C in numerator, e.g. 765 or
2 2 2 2 3
6×5×4 (e.g. 1 , 1 , 5 , 5 , 5 ) or 56
11 22 154 11 66 6
or diagram showing 6 blues and 7 gaps: M1
Numerator correct
Exact or 0.227 or better
Other methods: Correct answer B3, else 0
(e) | No because numerator and denominator would both be
multiplied by 5! [factor needn’t be present or correct]
or No as we are concerned only about the colour, OE | B1
[1] | 2.4 | No with relevant reason that is not wrong, e.g. “it
makes no difference that they now have labels”. More
than 1 reason: OK unless one is wrong
Not changed as you still treat them as yellow | B1
Different ways of arranging counters does not affect probability | B1
Not changed as being yellow is independent of numbers | B1
No as do not require a specific order of yellows or blues | B1
No as still 5 yellow counters to choose from | B1
No as it is only the colour that matters | B1
Both top and bottom are  same number so unchanged | B1
No as we use permutations not combinations | B1
No because still the same number of possibilities | B1
“No”, no reason | B0
Question | Answer | Marks | AO | Guidance
6 A bag contains 6 identical blue counters and 5 identical yellow counters.
\begin{enumerate}[label=(\alph*)]
\item Three counters are selected at random, without replacement.

Find the probability that at least two of the counters are blue.

All 11 counters are now arranged in a row in a random order.
\item Find the probability that all the yellow counters are next to each other.
\item Find the probability that no yellow counter is next to another yellow counter.
\item Find the probability that the counters are arranged in such a way that both of the following conditions hold.

\begin{itemize}
  \item Exactly three of the yellow counters are next to one another.
  \item Neither of the other two yellow counters is next to a yellow counter.
\item Explain whether the answer to part (d) would be different if the yellow counters were numbered $1,2,3,4$ and 5 , so that they are not identical.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2024 Q6 [11]}}
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