| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Standard +0.3 This is a standard S1 tree diagram question with conditional probabilities and a stopping condition. While it requires careful tracking of multiple paths and understanding of conditional probability, the structure is methodical and the calculations are straightforward. The multi-stage nature and independence assumption in later parts add some complexity, but this remains a typical textbook exercise requiring systematic application of learned techniques rather than novel insight. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct probabilities on tree diagram | G1 | Probabilities |
| Correct outcomes on tree diagram | G1 | Outcomes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{First team}) = 0.9^3 = 0.729\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{Second team}) = 0.9 \times 0.9 \times 0.1 + 0.9 \times 0.1 \times 0.5 + 0.1 \times 0.9 \times 0.5\) | M1 M1 | 1 correct triple; 3 correct triples added |
| \(= 0.081 + 0.045 + 0.045 = 0.171\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{asked to leave}) = 1 - 0.729 - 0.171 = 0.1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{Leave after two games} \mid \text{leaves}) = \frac{0.1 \times 0.5}{0.1} = \frac{1}{2}\) | M1 ft A1 cao | Denominator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{at least one asked to leave}) = 1 - 0.9^3 = 0.271\) | M1 ft M1 A1 cao | Calc'n of 0.9; \(1-(\ )^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{pass total of 7 games}) = P(\text{First, Second, Second}) + P(\text{First, First, Leave after three})\) | M1 | Attempts both |
| \(= 3 \times 0.729 \times 0.171^2 + 3 \times 0.729^2 \times 0.05\) | M1 ft M1 ft M1 | \(0.729(0.171)^2\); \(0.05(0.729)^2\); multiply by 3 |
| \(= 0.064 + 0.080 = 0.144\) | A1 cao |
## Question 6:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct probabilities on tree diagram | G1 | Probabilities |
| Correct outcomes on tree diagram | G1 | Outcomes |
### Part (ii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{First team}) = 0.9^3 = 0.729$ | A1 | |
### Part (ii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{Second team}) = 0.9 \times 0.9 \times 0.1 + 0.9 \times 0.1 \times 0.5 + 0.1 \times 0.9 \times 0.5$ | M1 M1 | 1 correct triple; 3 correct triples added |
| $= 0.081 + 0.045 + 0.045 = 0.171$ | A1 | |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{asked to leave}) = 1 - 0.729 - 0.171 = 0.1$ | B1 | |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{Leave after two games} \mid \text{leaves}) = \frac{0.1 \times 0.5}{0.1} = \frac{1}{2}$ | M1 ft A1 cao | Denominator |
### Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{at least one asked to leave}) = 1 - 0.9^3 = 0.271$ | M1 ft M1 A1 cao | Calc'n of 0.9; $1-(\ )^3$ |
### Part (vi)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{pass total of 7 games}) = P(\text{First, Second, Second}) + P(\text{First, First, Leave after three})$ | M1 | Attempts both |
| $= 3 \times 0.729 \times 0.171^2 + 3 \times 0.729^2 \times 0.05$ | M1 ft M1 ft M1 | $0.729(0.171)^2$; $0.05(0.729)^2$; multiply by 3 |
| $= 0.064 + 0.080 = 0.144$ | A1 cao | |
---6 Answer part (i) of this question on the insert provided.
Mancaster Hockey Club invite prospective new players to take part in a series of three trial games. At the end of each game the performance of each player is assessed as pass or fail. Players who achieve a pass in all three games are invited to join the first team squad. Players who achieve a pass in two games are invited to join the second team squad. Players who fail in two games are asked to leave. This may happen after two games.
\begin{itemize}
\item The probability of passing the first game is 0.9
\item Players who pass any game have probability 0.9 of passing the next game
\item Players who fail any game have probability 0.5 of failing the next game
\begin{enumerate}[label=(\roman*)]
\item On the insert, complete the tree diagram which illustrates the information above.\\
\includegraphics[max width=\textwidth, alt={}, center]{668963b4-994d-475a-a1c8-c3e3a252e4e6-4_691_1329_978_397}
\item Find the probability that a randomly selected player\\
(A) is invited to join the first team squad,\\
(B) is invited to join the second team squad.
\item Hence write down the probability that a randomly selected player is asked to leave.
\item Find the probability that a randomly selected player is asked to leave after two games, given that the player is asked to leave.
\end{itemize}
Angela, Bryony and Shareen attend the trials at the same time. Assuming their performances are independent, find the probability that
\item at least one of the three is asked to leave,
\item they pass a total of 7 games between them.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2005 Q6 [15]}}