OCR Further Statistics 2019 June — Question 3 6 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPermutations & Arrangements
TypeSpecific items separated
DifficultyChallenging +1.2 This is a Further Maths statistics question requiring systematic counting of arrangements with constraints. Part (a) demands careful case analysis of valid configurations with the 'two red counters minimum' constraint, while part (b) is more routine (stars and bars approach). The conceptual challenge and multi-step reasoning place it moderately above average difficulty.
Spec5.01a Permutations and combinations: evaluate probabilities
3 Six red counters and four blue counters are arranged in a straight line in a random order.
Find the probability that
  1. no blue counter has fewer than two red counters between it and the nearest other blue counter,
  2. no two blue counters are next to one another.
Question 3:
AnswerMarks Guidance
3(a) B R R B R R B R R B
4!6! 24720
  or 1 10!
6!4!
10! 3628800
 1 or 0.0047619…
AnswerMarks
210M1
M1
A1
AnswerMarks
[3]3.1b
1.1a
AnswerMarks
1.1This order inferred, e.g. diagram
(At least one ! or nC or nP)  10!
r r
Not  10C [except for 1  10C ]
4 4
AnswerMarks
Answer, exact or art 0.00476Alternatively:
4 65343221
10 9 8 7 6 5 4 3 2
M1A1 (1 error: M1A0)
Answer A1
AnswerMarks Guidance
(b) R
7C 6!4! 3572024
 4  or 7C  10!
10! 3628800 4 6!4!
 1 or 0.1666…
AnswerMarks
6M1
M1
A1
AnswerMarks
[3]3.1b
1.1a
AnswerMarks
1.1This structure implied, allow 6
7C or 7P or equivalent used
4 4
AnswerMarks
Answer, exact or 0.167 or betterMultiplication: M0M1A0
unless fully correct
Alternatively:
(5 + 20 + 10)/10C (any 3
4
terms correct): M2.  1 A1
6
Question 3:
3 | (a) | B R R B R R B R R B
4!6! 24720
  or 1 10!
6!4!
10! 3628800
 1 or 0.0047619…
210 | M1
M1
A1
[3] | 3.1b
1.1a
1.1 | This order inferred, e.g. diagram
(At least one ! or nC or nP)  10!
r r
Not  10C [except for 1  10C ]
4 4
Answer, exact or art 0.00476 | Alternatively:
4 65343221
10 9 8 7 6 5 4 3 2
M1A1 (1 error: M1A0)
Answer A1
(b) | | R | R | R | R | R | R | where the Bs go in a |
7C 6!4! 3572024
 4  or 7C  10!
10! 3628800 4 6!4!
 1 or 0.1666…
6 | M1
M1
A1
[3] | 3.1b
1.1a
1.1 | This structure implied, allow 6|
7C or 7P or equivalent used
4 4
Answer, exact or 0.167 or better | Multiplication: M0M1A0
unless fully correct
Alternatively:
(5 + 20 + 10)/10C (any 3
4
terms correct): M2.  1 A1
6
3 Six red counters and four blue counters are arranged in a straight line in a random order.\\
Find the probability that
\begin{enumerate}[label=(\alph*)]
\item no blue counter has fewer than two red counters between it and the nearest other blue counter,
\item no two blue counters are next to one another.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2019 Q3 [6]}}
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