OCR MEI Further Statistics A AS 2020 November — Question 3 8 marks

Exam BoardOCR MEI
ModuleFurther Statistics A AS (Further Statistics A AS)
Year2020
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeSecond success on trial n
DifficultyModerate -0.3 This is a straightforward application of geometric and binomial distribution formulas with clear parameters (p=12/52=3/13). Parts (a) and (c) are direct formula substitutions for geometric distribution, (b) uses complement rule, and (d) requires binomial calculation. All parts are standard textbook exercises requiring recall and basic computation rather than problem-solving insight.
Spec2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2

3 A child is trying to draw court cards from an ordinary pack of 52 cards (court cards are Kings, Queens and Jacks; there are 12 in a pack). She draws cards, one at a time, with replacement, from the pack. Find the probabilities of the following events.
  1. She draws a court card for the first time on the sixth try.
  2. She draws a court card at least once in the first six tries.
  3. She draws a court card for the second time on the sixth try.
  4. She draws at least two court cards in the first six tries.

Question 3:
AnswerMarks Guidance
3(a) P(Sixth try) =
5
10 3
= 0.0622
AnswerMarks
�13� �13�M1
A1
AnswerMarks
[2]3.3
1.1
AnswerMarks Guidance
3(b) P(at least once in first six) =
6
= 0.792180
AnswerMarks
1−�13�M1
A1
AnswerMarks
[2]1.1a
1.1
AnswerMarks Guidance
3(c) P(second time sixth try) =
4 2
= 0.093120 3
AnswerMarks
5�13� �13�M1
A1
AnswerMarks Guidance
[2]3.1b
1.1Allow M1 if 5 omitted
3(d) P(at least 2) = 1 – 0.5801
= 0.4199M1
A1
AnswerMarks
[2]3.1b
1.1M1 for 1 – P(X = 0 or 1) using
B(6, 3/13)
BC
Question 3:
3 | (a) | P(Sixth try) =
5
10 3
= 0.0622
�13� �13� | M1
A1
[2] | 3.3
1.1
3 | (b) | P(at least once in first six) =
6
= 0.792180
1−�13� | M1
A1
[2] | 1.1a
1.1
3 | (c) | P(second time sixth try) =
4 2
= 0.093120 3
5�13� �13� | M1
A1
[2] | 3.1b
1.1 | Allow M1 if 5 omitted
3 | (d) | P(at least 2) = 1 – 0.5801
= 0.4199 | M1
A1
[2] | 3.1b
1.1 | M1 for 1 – P(X = 0 or 1) using
B(6, 3/13)
BC
3 A child is trying to draw court cards from an ordinary pack of 52 cards (court cards are Kings, Queens and Jacks; there are 12 in a pack). She draws cards, one at a time, with replacement, from the pack.

Find the probabilities of the following events.
\begin{enumerate}[label=(\alph*)]
\item She draws a court card for the first time on the sixth try.
\item She draws a court card at least once in the first six tries.
\item She draws a court card for the second time on the sixth try.
\item She draws at least two court cards in the first six tries.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2020 Q3 [8]}}