| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Mean/expectation of geometric distribution |
| Difficulty | Standard +0.3 Part (i) involves direct application of standard geometric distribution formulas (expectation, PMF, and tail probability) requiring only recall and basic calculation. Part (ii) requires recognizing that P(Y is odd) forms a geometric series and applying the sum formula—a modest step up requiring some algebraic manipulation but still a standard textbook exercise with clear scaffolding. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks |
|---|---|
| \(= 5\) | M1 |
| A1 | .2.. |
| Answer | Marks |
|---|---|
| \((^1/_5)^5 × ^1/_5 = a/_{625}\) or \(0.102\) (3 sfs) | M1 |
| A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((^1/_5)^3\) | M1 | or \(1-(^1/_5 + ^4/_5×^1/_5 + (^1/_5)^2×^1/_5)\) |
| NOT \(1 - (^1/_5)^4\) | ||
| \(= \frac{256}{_{655}}\) or a.r.t \(0.410\) (3 sfs) or \(0.41\) | A1 | .2.. |
| Answer | Marks | Guidance |
|---|---|---|
| \(p, p(1 - p)^2, p(1 - p)^4\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Or give indication of how terms derived | M1 | ≥ two terms |
| Answer | Marks | Guidance |
|---|---|---|
| Recog that c.r.t. \(= q^2\) or \((1-p)^2\) | M1 | or eg \(r = q^2p/p\) |
| \(S_∞ = \frac{p}{1-q^2}\) or \(\frac{p}{1-(1-p)^2}\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1-q}{(1-q)(1+q)}\) | M1 | ( \(= \frac{p}{2p - p^2}\) ) \(= \frac{p}{p(2-p)}\) |
| ( \(= \frac{1}{2 - p}\) ) \(= \frac{1}{2-(1-q)}\) | ||
| Must see this step for A1 | ||
| A1 | 4 |
### Part ia
$1 / ^1/_5$
$= 5$ | M1 |
| A1 | .2..
### Part b
$(^1/_5)^5 × ^1/_5 = a/_{625}$ or $0.102$ (3 sfs) | M1 |
| A1 | 2
### Part c
$(^1/_5)^3$ | M1 | or $1-(^1/_5 + ^4/_5×^1/_5 + (^1/_5)^2×^1/_5)$
| | NOT $1 - (^1/_5)^4$
$= \frac{256}{_{655}}$ or a.r.t $0.410$ (3 sfs) or $0.41$ | A1 | .2..
### Part iii
$P(Y=1) = p, P(Y=3) = q^2p, P(Y=5) = q^4p$
$p, p(1 - p)^2, p(1 - p)^4$ | B1 | 1
$q^4, q^5, q^6$
or any of these with $1 - p$ instead of $q$
"Always $q$ to even power × $p$"
Either associate each term with relevant prob
Or give indication of how terms derived | M1 | ≥ two terms
### Part b
Recog that c.r.t. $= q^2$ or $(1-p)^2$ | M1 | or eg $r = q^2p/p$
$S_∞ = \frac{p}{1-q^2}$ or $\frac{p}{1-(1-p)^2}$ | M1 |
$P(\text{odd}) = \frac{1-q}{1-q^2}$
$= \frac{1-q}{(1-q)(1+q)}$ | M1 | ( $= \frac{p}{2p - p^2}$ ) $= \frac{p}{p(2-p)}$
| | ( $= \frac{1}{2 - p}$ ) $= \frac{1}{2-(1-q)}$
| | Must see this step for A1
| A1 | 4
**Total: 11**9 (i) A random variable $X$ has the distribution $\operatorname { Geo } \left( \frac { 1 } { 5 } \right)$. Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { E } ( \mathrm { X } )$,
\item $\mathrm { P } ( \mathrm { X } = 4 )$,
\item $P ( X > 4 )$.\\
(ii) A random variable $Y$ has the distribution $\operatorname { Geo } ( p )$, and $q = 1 - p$.\\
(a) Show that $P ( Y$ is odd $) = p + q ^ { 2 } p + q ^ { 4 } p + \ldots$.\\
(b) Use the formula for the sum to infinity of a geometric progression to show that
$$P ( Y \text { is odd } ) = \frac { 1 } { 1 + q }$$
{}\\
7
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2007 Q9 [11]}}