OCR MEI S1 2013 January — Question 5 5 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2013
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeFirst success on specific trial
DifficultyEasy -1.2 This is a straightforward application of the geometric distribution formula with no conceptual challenges. Part (i) requires a single calculation using P(X=3) = (5/6)²(1/6), and part (ii) uses the complement rule P(X≤10) = 1-(5/6)¹⁰. Both are standard textbook exercises requiring only direct formula application with no problem-solving or insight needed.
Spec5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)

5 Malik is playing a game in which he has to throw a 6 on a fair six-sided die to start the game. Find the probability that
  1. Malik throws a 6 for the first time on his third attempt,
  2. Malik needs at most ten attempts to throw a 6.

Part (i)
Answer: \(\left(\frac{5}{6}\right)^2 \times \frac{1}{6} = \frac{25}{216} (= 0.116)\)
AnswerMarks
MarksGuidance
M1For 5/6 (or \(1 - 1/6\)) seen
M1For whole product
A1cao
[3]Allow 0.12 with working. If extra term or whole number factor present give M1M0A0
Part (ii)
Answer: \(1 - \left(\frac{5}{6}\right)^{10} = 1 - 0.1615 = 0.8385\)
AnswerMarks
MarksGuidance
M1For \((5/6)^{10}\) (without extra terms)
A1cao
[2]Allow 0.838 or 0.839 without working and 0.84 with working. For addition \(P(X=1) + \ldots + P(X=10)\) give M1A1 for 0.84 or better, otherwise M0A0
## Part (i)

**Answer:** $\left(\frac{5}{6}\right)^2 \times \frac{1}{6} = \frac{25}{216} (= 0.116)$

| **Marks** | **Guidance** |
|-----------|--------------|
| M1 | For 5/6 (or $1 - 1/6$) seen |
| M1 | For whole product |
| A1 | cao |
| **[3]** | Allow 0.12 with working. If extra term or whole number factor present give M1M0A0 |

## Part (ii)

**Answer:** $1 - \left(\frac{5}{6}\right)^{10} = 1 - 0.1615 = 0.8385$

| **Marks** | **Guidance** |
|-----------|--------------|
| M1 | For $(5/6)^{10}$ (without extra terms) |
| A1 | cao |
| **[2]** | Allow 0.838 or 0.839 without working and 0.84 with working. For addition $P(X=1) + \ldots + P(X=10)$ give M1A1 for 0.84 or better, otherwise M0A0 |

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5 Malik is playing a game in which he has to throw a 6 on a fair six-sided die to start the game. Find the probability that\\
(i) Malik throws a 6 for the first time on his third attempt,\\
(ii) Malik needs at most ten attempts to throw a 6.

\hfill \mbox{\textit{OCR MEI S1 2013 Q5 [5]}}