| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Moderate -0.3 This is a straightforward conditional probability question using tree diagrams and expectation. Part (i) involves routine probability calculations without replacement (19/20, 18/19, etc.) and summing paths. Part (ii) requires finding a probability distribution and calculating E(X), which are standard S1 techniques with no conceptual challenges. The arithmetic is slightly tedious but the methods are entirely routine for this module. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
### Part ia
**Answer:** $\frac{15}{19}$ or $\frac{1}{19}$ seen or $\frac{1}{18}$ or $\frac{1}{8}$ seen. structure correct to 6 branches. all correct incl. probs and W & R
**Marks:** B1, B1, B1, B1 4
**Guidance:** regardless of probs & labels (or 14 branches with correct 0s & 1s)
### Part b
**Answer:** $\frac{1}{20} + ^{19}/_{20} × \frac{1}{19} + ^{19}/_{20} × ^{18}/_{19} × \frac{1}{18}$ = $\frac{1}{20}$
**Marks:** M2, A1 3
**Guidance:** M1 any 2 correct $^{19}/_{20} × \frac{1}{19}$ terms added. or $\frac{^{19}/_{20} × ^{19}/_{20} × ^{17}/_{18}}{1 - ^{19}/_{20} × \frac{1}{19} × ^{17}_{20}}$ or $1 - (1 - \frac{1}{20}) × (^{19}/_{20})$
### Part iia
**Answer:** $\frac{^{19}/_{20} × ^{18}/_{19}}{= \frac{1}{10}}$ oe
**Marks:** M1, A1 2
**Guidance:** $\frac{^{19}/_{20} × ^{18}/_{19} × ^{17}/_{18} + \frac{^{19}/_{20} × ^{18}/_{19} × ^{17}/_{18} × \frac{1}{17}}{= \frac{1}{10}}$ or $\frac{^{19}/_{20} + \frac{1}{20}$
### Part b
**Answer:** (P(X = 1) = $\frac{1}{20}$) or $\frac{1}{20} × \frac{1}{19} = \frac{1}{20}$. $\Sigma xp = ^{19}/_{20}$ or 2.85
**Marks:** M1, A1 1, A1 4
**Guidance:** $\geq 2$ terms, fit their p's if $\Sigma p = 1$. NB: $^{19}/_{20}×3 = 2.85$ no mks
### Parts ia, ib, iia, iib
**Answer:** With replacement: Original scheme: $\frac{1}{20} + ^{19}/_{20} × \frac{1}{20} + (^{20}/_{20})^2 × (^{1}/_{20}) × ^{20}/_{20}$ or $1 - (^{19}/_{20})^2$. or $1 - (^{19}/_{20})$ or 2 probs of $\frac{1}{20}$ M1A1. Original scheme: But NB ans 2.85(25...) M1A0M1A0
**Marks:** M1A0M1A0
---A game at a charity event uses a bag containing 19 white counters and 1 red counter. To play the game once a player takes counters at random from the bag, one at a time, without replacement. If the red counter is taken, the player wins a prize and the game ends. If not, the game ends when 3 white counters have been taken. Niko plays the game once.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Copy and complete the tree diagram showing the probabilities for Niko. [4]
\includegraphics{figure_2}
\item Find the probability that Niko will win a prize. [3]
\end{enumerate}
\item The number of counters that Niko takes is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item Find P($X = 3$). [2]
\item Find E($X$). [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2009 Q8 [13]}}