| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric and negative binomial distributions with standard probability calculations. Parts (i)-(iv) are routine textbook exercises requiring direct formula application, while part (v) involves recognizing that the sum of two geometric distributions gives the expected value, which is slightly above pure recall but still standard Further Maths content. |
| Spec | 2.04b 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:169)6(cid:185) 6 | B1 | |
| [1] | 1.1 | E |
| 4 | (ii) | 4 |
| Answer | Marks |
|---|---|
| =0.518 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | M | |
| 4 | (iii) | 2 |
| Answer | Marks |
|---|---|
| 216 108 | E |
| Answer | Marks |
|---|---|
| [2] | C |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (iv) | 2 |
| Answer | Marks |
|---|---|
| 108 27 | M1 |
| A1 | 3.1b |
| Answer | Marks |
|---|---|
| Alternative Method | N |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:169)6(cid:185) 6 (cid:169)6(cid:185) | M1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 216 216 27 | A1 | E |
| [2] | M | |
| 4 | (v) | Expected value for one five = 6 |
| Answer | Marks |
|---|---|
| So for two fives expected value = 6 + 6 = 12 | E1 |
| Answer | Marks |
|---|---|
| [3] | I |
| Answer | Marks |
|---|---|
| 2.1 | soi |
Question 4:
4 | (i) | 3
(cid:167)5(cid:183) 1
(cid:168) (cid:184) (cid:117) (cid:32)0.0965
(cid:169)6(cid:185) 6 | B1
[1] | 1.1 | E
4 | (ii) | 4
(cid:167)5(cid:183)
1(cid:16)(cid:168) (cid:184)
(cid:169)6(cid:185)
=0.518 | M1
A1
[2] | 1.1a
I
1.1 | M
4 | (iii) | 2
5 (cid:167)1(cid:183) 1 5 1
(cid:117)(cid:168) (cid:184) (cid:14) (cid:117) (cid:117)
6 (cid:169)6(cid:185) 6 6 6
P
10 5
(cid:32) (cid:32) (cid:32)0.0463
216 108 | E
M1
A1
[2] | C
3.1b
1.1
4 | (iv) | 2
5 (cid:167)1(cid:183)
(cid:14)(cid:168) (cid:184)
108 (cid:169)6(cid:185)
8 2
(cid:32) (cid:32) (cid:32)0.0741
108 27 | M1
A1 | 3.1b
1.1
Alternative Method | N
2 3
(cid:167)1(cid:183) 5 (cid:167)1(cid:183)
3(cid:117)(cid:168) (cid:184) (cid:117) (cid:14)(cid:168) (cid:184)
(cid:169)6(cid:185) 6 (cid:169)6(cid:185) | M1 | M1
15 1 2
(cid:32) (cid:14) (cid:32) (cid:32)0.0741
216 216 27 | A1 | E
[2] | M
4 | (v) | Expected value for one five = 6
Because geometric
So for two fives expected value = 6 + 6 = 12 | E1
E1
B1
[3] | I
1.1
C
2.4
2.1 | soi4 A fair six-sided dice is rolled repeatedly. Find the probability of the following events.\\
(i) A five occurs for the first time on the fourth roll.\\
(ii) A five occurs at least once in the first four rolls.\\
(iii) A five occurs for the second time on the third roll.\\
(iv) At least two fives occur in the first three rolls.
The dice is rolled repeatedly until a five occurs for the second time.\\
(v) Find the expected number of rolls required for two fives to occur. Justify your answer.
\hfill \mbox{\textit{OCR MEI Further Statistics Major Q4 [10]}}