| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Specific items separated |
| Difficulty | Challenging +1.2 This is a multi-part combinatorics problem requiring systematic case analysis and careful counting. Part (a) involves straightforward hypergeometric probability with 'at least' requiring summing cases. Part (b)(i) uses the standard 'treat as a block' technique with verification. Part (b)(ii) requires more sophisticated reasoning to avoid overcounting when splitting vowels into two groups. While methodical, it demands careful organization and is more involved than typical A-level questions, placing it moderately above average difficulty for Further Maths statistics. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 5C ×21C +5C ×21C +1 [= 2100 + 105 + 1] |
| Answer | Marks |
|---|---|
| 32890 | M1dep |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | Any correct pair of nCs multiplied |
| Answer | Marks |
|---|---|
| Awrt 0.0335 or any exact fraction | Or 1 – P(0, 1, 2) = 1 – .9665 |
| Answer | Marks |
|---|---|
| OR: | Or: 5 × 4 × 3 × 2 × 1 |
Total 1103 or 0.0335…
| Answer | Marks | Guidance |
|---|---|---|
| 32890 | B1 | Must have 5 oe, e.g. 5C |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | (i) | 22!×5! 1×2×3×4×5 120 |
| Answer | Marks |
|---|---|
| 2990 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2a | Oe. Allow M1 for 21! instead of 22! |
| Answer | Marks |
|---|---|
| method | 1×2×3×4×5 |
| Answer | Marks |
|---|---|
| (ii) | 22 fences: 22 for [VVV] × 21 for [VV] |
| Answer | Marks |
|---|---|
| (= 2.832×1024 ÷ 4.0329×1026) | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | Correct strategy, allow 22C for 22P |
| Answer | Marks |
|---|---|
| exact fraction | 21!×3!×2!×22×21: M2A0 |
| Answer | Marks | Guidance |
|---|---|---|
| OR: | Treat 21 consonants, [VVV] and [VV] as 23 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | Correct strategy, allow 23!×2!×3! | (Must subtract 2 × 1/2990 as |
| 23! × 5! / 26! (= 1/130) | Correct (5! = 5P ×2P = 5C ×2!×3!) | |
| 3 2 3 | 23! method counts | |
| Subtract 2 × 1/2990 | M1 also for subtracting 1 × 1/2990 | [VVVVV] twice, once |
| Answer | Marks | Guidance |
|---|---|---|
| 2990 | Final answer, exact fraction | as [VVV][VV] and once |
| (11/1495 is M1A1M1A0) | as [VV][VVV]) |
Total 1103 or 0.0335…
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | (a) | 5C ×21C +5C ×21C +1 [= 2100 + 105 + 1]
3 2 4 1
÷ 26C [= 65780]
5
1103 or 0.0335…
32890 | M1dep
A1
*depM1
A1
[4] | 3.1b
1.1
1.1
3.2a | Any correct pair of nCs multiplied
r
All terms correct
Awrt 0.0335 or any exact fraction | Or 1 – P(0, 1, 2) = 1 – .9665
2206 264720
e.g. or
65780 7893600
OR: | Or: 5 × 4 × 3 × 2 × 1
26 25 24 23 22
5 × 4 × 3 × 2 ×21×5
26 25 24 23 22
5 × 4 × 3 ×21×20×10
26 25 24 23 22
Total 1103 or 0.0335…
32890 | B1 | Must have 5 oe, e.g. 5C
1
Must have 10 oe, e.g. 5C
3
B1
B1
B1
[4]
(b) | (i) | 22!×5! 1×2×3×4×5 120
(= = )
26! 23×24×25×26 358800
1
= AG
2990 | M1
A1
A1
[3] | 1.1
2.1
2.2a | Oe. Allow M1 for 21! instead of 22!
Fully correct
Correctly obtain AG using exact
method | 1×2×3×4×5
: M1
22×23×24×25×26
Allow even if no working
after 22! × 5! ÷ 26!
(ii) | 22 fences: 22 for [VVV] × 21 for [VV]
Consonants arranged in 21! ways
Vowels arranged in 5! ways (= 5P × 2P )
3 2
21
Product ÷ 26! =
2990
(= 2.832×1024 ÷ 4.0329×1026) | M1
M1
A1
A1
[4] | 3.1b
1.1
2.1
3.2a | Correct strategy, allow 22C for 22P
2 2
At least one of these, no subtraction
Both correct
Allow from calculator but must be
exact fraction | 21!×3!×2!×22×21: M2A0
21! × 3! × 2! ÷ 26! M0M1
5C × 3! × 2! = 5!
3
OR: | Treat 21 consonants, [VVV] and [VV] as 23 | M1
A1
M1
A1
[4] | 3.1b
2.1
3.2a
1.1 | Correct strategy, allow 23!×2!×3! | (Must subtract 2 × 1/2990 as
23! × 5! / 26! (= 1/130) | Correct (5! = 5P ×2P = 5C ×2!×3!)
3 2 3 | 23! method counts
Subtract 2 × 1/2990 | M1 also for subtracting 1 × 1/2990 | [VVVVV] twice, once
21
Answer is
2990 | Final answer, exact fraction | as [VVV][VV] and once
(11/1495 is M1A1M1A0) | as [VV][VVV])
Or: 5 × 4 × 3 × 2 × 1
26 25 24 23 22
5 × 4 × 3 × 2 ×21×5
26 25 24 23 22
5 × 4 × 3 ×21×20×10
26 25 24 23 22
Total 1103 or 0.0335…
32890
Must have 5 oe, e.g. 5C
1
Must have 10 oe, e.g. 5C
3
M1
A1
M1
A1
3.1b
2.1
3.2a
1.1
Question | Answer | Marks | AO | Guidance26 cards are each labelled with a different letter of the alphabet, A to Z. The letters A, E, I, O and U are vowels.
\begin{enumerate}[label=(\alph*)]
\item Five cards are selected at random without replacement.
Determine the probability that the letters on at least three of the cards are vowels. [4]
\item All 26 cards are arranged in a line, in random order.
\begin{enumerate}[label=(\roman*)]
\item Show that the probability that all the vowels are next to one another is $\frac{1}{2990}$. [3]
\item Determine the probability that three of the vowels are next to each other, and the other two vowels are next to each other, but the five vowels are not all next to each other. [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2020 Q5 [11]}}