| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Second success on trial n |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric and negative binomial distributions with clearly defined scenarios. Part (i) requires basic geometric distribution calculations (probability of first success on 4th trial, and cumulative probability), while part (ii) involves negative binomial distribution for two successes. All parts follow standard textbook patterns with no conceptual tricks, though the negative binomial in part (ii) requires slightly more careful calculation, making it slightly easier than average overall. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Geo(0.3) stated or implied, 0.7³ × 0.3 = 0.103 (3 sf) | M1, M1, A1 3 | by 0.7³ × 0.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: 0.7³ or 0.343, \(1-0.7^3\) | M1, M1 | 0.7³ must be alone, ie not 0.7³ × 0.3 or similar, allow 1 - 0.74 or 0.7599 or 0.76 for M1 only |
| or \(0.3 + 0.7×0.3 + 0.7^2×0.3\); 1 term wrong or omitted or extra or \(1-(0.3+0.7×0.3+0.7^2×0.3)\) or 0.343; | M1M1, M1 | |
| = 0.657 | A1 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: State or imply one viewer in 1st four | M1 | or B(4, 0.3) stated, or \(^4C_1\) used, or YNNNY |
| \(^4C_1×0.7^3×0.3\) (= 0.412) | M1, M1 | dep 1st M1 |
| \(×0.3\) | M1 | |
| = 0.123 (3 sf) | A1 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(0.7^3 + ^4C_1×0.74×0.3\) = 0.528 (3 sf) | M1, A1 2 | or \(1-(0.3^2+2×0.3×0.7×0.7+3×0.3×0.7+4×0.3×0.7)\) Not ISW, eg 1 - 0.528; M1A0 |
## Part ia:
**Answer/Working:** Geo(0.3) stated or implied, 0.7³ × 0.3 = 0.103 (3 sf) | **M1, M1, A1 3** | by 0.7³ × 0.3
## Part b:
**Answer/Working:** 0.7³ or 0.343, $1-0.7^3$ | **M1, M1** | 0.7³ must be alone, ie not 0.7³ × 0.3 or similar, allow 1 - 0.74 or 0.7599 or 0.76 for M1 only
or $0.3 + 0.7×0.3 + 0.7^2×0.3$; 1 term wrong or omitted or extra or $1-(0.3+0.7×0.3+0.7^2×0.3)$ or 0.343; | **M1M1, M1** |
= 0.657 | **A1 3** |
## Part iii a:
**Answer/Working:** State or imply one viewer in 1st four | **M1** | or B(4, 0.3) stated, or $^4C_1$ used, or YNNNY
$^4C_1×0.7^3×0.3$ (= 0.412) | **M1, M1** | dep 1st M1
$×0.3$ | **M1** |
= 0.123 (3 sf) | **A1 4** |
## Part b:
**Answer/Working:** $0.7^3 + ^4C_1×0.74×0.3$ = 0.528 (3 sf) | **M1, A1 2** | or $1-(0.3^2+2×0.3×0.7×0.7+3×0.3×0.7+4×0.3×0.7)$ Not ISW, eg 1 - 0.528; M1A0
**Total: 12 marks**
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**Total: 72 marks**The proportion of people who watch West Street on television is 30\%. A market researcher interviews people at random in order to contact viewers of West Street. Each day she has to contact a certain number of viewers of West Street.
\begin{enumerate}[label=(\roman*)]
\item Near the end of one day she finds that she needs to contact just one more viewer of West Street. Find the probability that the number of further interviews required is
\begin{enumerate}[label=(\alph*)]
\item 4, [3]
\item less than 4. [3]
\end{enumerate}
\item Near the end of another day she finds that she needs to contact just two more viewers of West Street. Find the probability that the number of further interviews required is
\begin{enumerate}[label=(\alph*)]
\item 5, [4]
\item more than 5. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2010 Q8 [12]}}