OCR MEI Paper 2 2022 June — Question 11 10 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeProbabilities in table form with k
DifficultyStandard +0.3 This is a straightforward probability distribution question requiring basic probability axioms (probabilities sum to 1), expected value calculation, and a simple binomial probability. All parts follow standard textbook procedures with no novel insight required, making it slightly easier than average for A-level.
Spec2.02f Measures of average and spread2.04a Discrete probability distributions
A die in the form of a dodecahedron has its faces numbered from 1 to 12. The die is biased so that the probability that a score of 12 is achieved is different from any other score. The probability distribution of \(X\), the score on the die, is given in the table in terms of \(p\) and \(k\), where \(0 < p < 1\) and \(k\) is a positive integer.
\(x\)123456789101112
P\((X = x)\)\(p\)\(p\)\(p\)\(p\)\(p\)\(p\)\(p\)\(p\)\(p\)\(p\)\(p\)\(kp\)
Sam rolls the die 30 times, Leo rolls the die 60 times and Nina rolls the die 120 times. They each plot their scores on bar line graphs.
  1. Explain whose graph is most likely to give the best representation of the theoretical probability distribution for the score on the die. [1]
  2. Find \(p\) in terms of \(k\). [2]
  3. Determine, in terms of \(k\), the expected number of times Nina rolls a 12. [3]
  4. Given that Nina rolls a 12 on 32 occasions, calculate an estimate of the value of \(k\). [2]
Nina rolls the die a further 30 times.
  1. Use your answer to part (d) to calculate an estimate for the probability that she obtains a 12 exactly 8 times in these 30 rolls. [2]
Question 11:
AnswerMarks Guidance
11(a) Nina’s, because hers is the largest sample
size oeB1 2.2a
probabilities get closer to the true probabilities oe
[1]
AnswerMarks Guidance
11(b) 11p + kp =1
1
p =
AnswerMarks
11+π‘˜M1
A13.1a
1.1
[2]
AnswerMarks Guidance
11(c) 1 1
π‘‘β„Žπ‘’π‘–π‘Ÿ Γ— π‘˜ or π‘‘β„Žπ‘’π‘–π‘Ÿ Γ—120
11+π‘˜ 11+π‘˜
π‘˜
120Γ—π‘‘β„Žπ‘’π‘–π‘Ÿ
11+π‘˜
120π‘˜
oe
AnswerMarks
11+π‘˜M1
M1
AnswerMarks
A12.1
1.2
AnswerMarks
1.1multiply by k or by 120; may be embedded
multiplying by both k and 120
[3]
AnswerMarks Guidance
11(d) 120π‘˜
32 = their oe
11+π‘˜
AnswerMarks
k = 4M1
A11.1
1.1
[2]
Alternatively
32 1
11𝑝 = 1βˆ’ may be implied by 𝑝 =
120 15
(from (𝑃(𝑋 β‰  12))
AnswerMarks
k = 4π‘˜π‘“ 32
or =
120 120
(from 11f = 120 β€’ 32 = 88 so f = 8 and so π‘˜π‘ = β‹―)
k = 4
M1
A1
AnswerMarks Guidance
11(e) 4 32
π‘Œ~B(30,π‘‘β„Žπ‘’π‘–π‘Ÿ ) or π‘Œ~B(30, ) used
11+4 120
to find P(Y = 8)
AnswerMarks
0.16 – 0.163 BCM1
A13.1a
1.1Y is the number of 12s obtained in 30 rolls;
allow B2 for 0.1628 – 0.163 unsupported
[2]
Question 11:
11 | (a) | Nina’s, because hers is the largest sample
size oe | B1 | 2.2a | allow eg Nina’s, because with a larger sample size the
probabilities get closer to the true probabilities oe
[1]
11 | (b) | 11p + kp =1
1
p =
11+π‘˜ | M1
A1 | 3.1a
1.1
[2]
11 | (c) | 1 1
π‘‘β„Žπ‘’π‘–π‘Ÿ Γ— π‘˜ or π‘‘β„Žπ‘’π‘–π‘Ÿ Γ—120
11+π‘˜ 11+π‘˜
π‘˜
120Γ—π‘‘β„Žπ‘’π‘–π‘Ÿ
11+π‘˜
120π‘˜
oe
11+π‘˜ | M1
M1
A1 | 2.1
1.2
1.1 | multiply by k or by 120; may be embedded
multiplying by both k and 120
[3]
11 | (d) | 120π‘˜
32 = their oe
11+π‘˜
k = 4 | M1
A1 | 1.1
1.1
[2]
Alternatively
32 1
11𝑝 = 1βˆ’ may be implied by 𝑝 =
120 15
(from (𝑃(𝑋 β‰  12))
k = 4 | π‘˜π‘“ 32
or =
120 120
(from 11f = 120 β€’ 32 = 88 so f = 8 and so π‘˜π‘ = β‹―)
k = 4
M1
A1
11 | (e) | 4 32
π‘Œ~B(30,π‘‘β„Žπ‘’π‘–π‘Ÿ ) or π‘Œ~B(30, ) used
11+4 120
to find P(Y = 8)
0.16 – 0.163 BC | M1
A1 | 3.1a
1.1 | Y is the number of 12s obtained in 30 rolls;
allow B2 for 0.1628 – 0.163 unsupported
[2]
A die in the form of a dodecahedron has its faces numbered from 1 to 12. The die is biased so that the probability that a score of 12 is achieved is different from any other score. The probability distribution of $X$, the score on the die, is given in the table in terms of $p$ and $k$, where $0 < p < 1$ and $k$ is a positive integer.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
P$(X = x)$ & $p$ & $p$ & $p$ & $p$ & $p$ & $p$ & $p$ & $p$ & $p$ & $p$ & $p$ & $kp$ \\
\hline
\end{tabular}
\end{center}

Sam rolls the die 30 times, Leo rolls the die 60 times and Nina rolls the die 120 times. They each plot their scores on bar line graphs.

\begin{enumerate}[label=(\alph*)]
\item Explain whose graph is most likely to give the best representation of the theoretical probability distribution for the score on the die. [1]
\item Find $p$ in terms of $k$. [2]
\item Determine, in terms of $k$, the expected number of times Nina rolls a 12. [3]
\item Given that Nina rolls a 12 on 32 occasions, calculate an estimate of the value of $k$. [2]
\end{enumerate}

Nina rolls the die a further 30 times.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Use your answer to part (d) to calculate an estimate for the probability that she obtains a 12 exactly 8 times in these 30 rolls. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Paper 2 2022 Q11 [10]}}
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