| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Sequential dependent events |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question using basic counting principles. Parts (i)(a-c) require simple fraction calculations with reduced sample spaces, and part (ii) tests recognition that sampling without replacement violates the independence requirement for geometric distributions. All steps are routine applications of S1 concepts with no problem-solving insight needed. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles5.02f Geometric distribution: conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \(^1/_3\) oe | B1 | 1 |
| \(B↔W\) MR: max (a)B0(b)M1(c)B1M1(c)B1M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= ^1/_10 × ^1/_9 + ^9/_10 × ^1/_9\) or \(^1/_15 + ^1/_15 = ^2/_15\) oe | M1 | Or \(^1/_10 × ^2/_9\) OR \(^6/_10 × ^2/_9\) correct |
| M1 | NB \(^8/_10 × ^1/_10\); \(^6/_10 × ^1/_10 = ^2/_15\): M1M0A0 | |
| A1 | ||
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(^3/_9 × ^2/_8 + ^9/_9 × ^8/_8\) | B1 | Or \(^2/_15\) as numerator |
| M1 | Or \(\frac{^1/_9×^2/_8+^0/_9×^2}_8}{^1/_9×^2/_8+^0/_9×^2}_8}\) | |
| above \(+ ^{}_10×^{}/_9×^8/_8 + ^0/_10×^7/_9×^{}_8\) | ||
| A1 | May not see wking | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\)Blue\()\) not constant or discs not indep, so no | B1 | 1 |
| Prob changes as discs removed | ||
| Limit to no. of discs. Fixed no. of discs | ||
| Discs will run out | ||
| Context essential: "disc" or "blue" | ||
| NOT fixed no. of trials | ||
| NOT because without repl Ignore extra |
### Part ia
$^1/_3$ oe | B1 | 1
| | $B↔W$ MR: max (a)B0(b)M1(c)B1M1(c)B1M1
### Part b
$P(BB) + P(WB)$ attempted
$= ^1/_10 × ^1/_9 + ^9/_10 × ^1/_9$ or $^1/_15 + ^1/_15 = ^2/_15$ oe | M1 | Or $^1/_10 × ^2/_9$ OR $^6/_10 × ^2/_9$ correct
| M1 | NB $^8/_10 × ^1/_10$; $^6/_10 × ^1/_10 = ^2/_15$: M1M0A0
| A1 |
| 5 |
### Part c
Dedoms 9 & 8 seen or implied
$^3/_9 × ^2/_8 + ^9/_9 × ^8/_8$ | B1 | Or $^2/_15$ as numerator
| M1 | Or $\frac{^1/_9×^2/_8+^0/_9×^2}_8}{^1/_9×^2/_8+^0/_9×^2}_8}$
| | above $+ ^{}_10×^{}/_9×^8/_8 + ^0/_10×^7/_9×^{}_8$
| A1 | May not see wking
| 2 |
### Part ii
$P($Blue$)$ not constant or discs not indep, so no | B1 | 1
| | Prob changes as discs removed
| | Limit to no. of discs. Fixed no. of discs
| | Discs will run out
| | Context essential: "disc" or "blue"
| | NOT fixed no. of trials
| | NOT because without repl Ignore extra
**Total: 8**
---4 A bag contains 6 white discs and 4 blue discs. Discs are removed at random, one at a time, without replacement.\\
(i) Find the probability that
\begin{enumerate}[label=(\alph*)]
\item the second disc is blue, given that the first disc was blue,
\item the second disc is blue,
\item the third disc is blue, given that the first disc was blue.\\
(ii) The random variable $X$ is the number of discs which are removed up to and including the first blue disc. State whether the variable X has a geometric distribution. Explain your answer briefly.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2007 Q4 [8]}}