| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Repeated trials until stopping condition |
| Difficulty | Standard +0.3 This is a standard tree diagram question with independent trials and a stopping condition. Parts (a)-(d) require systematic enumeration of outcomes using basic probability rules (multiplication and addition). Part (e) involves conditional probability P(A|B) = P(A∩B)/P(B), which is routine at A-level. The question is slightly easier than average because the structure is clearly guided across multiple parts, the probabilities are simple decimals, and tree diagrams for 'first to k wins' are a common textbook exercise. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(0\) | B1 [1] | Allow 0% |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram with correct structure: \((A)\) branches 0.7, 0.3 leading to \(A\)/\(B\); \((0.7)\) branches 0.7, 0.3; \((0.3)\) branches 0.7, 0.3 leading to \(A\)/\(B\) | B1, B1, B1 [3] | B2: 8 correct branches and probs OR names, no extra branches. B2: 7 correct branches, probs and names, no extra branches. B1: 8 correct branches without probs & names. B1: 6 correct branches, probs and names. Ignore products at ends. Ignore extra branches if no probabilities or \(p=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.3 \times 0.3 + 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.3\) or \(0.09 + 0.063 + 0.063\) | M1 | All correct; M2 ft their diagram |
| Two products correct | M1 | ft their diagram |
| \(= \frac{27}{125}\) or \(0.216\) | A1 [3] | SC Correct answer with no working: B2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.7 \times 0.3 + 0.3 \times 0.7\) or \(0.21 + 0.21\) | M1 | or \(1-(0.7^2+0.3^2)\) or \(1-(0.49+0.09)\). Condone missing brackets. Or \(0.7\times0.3\times0.7 + 0.7\times0.3\times0.3 + 0.3\times0.7\times0.7 + 0.3\times0.7\times0.3\) or \(2\times0.147 + 2\times0.063\) |
| \(= \frac{21}{50}\) or \(0.42\) | A1 [2] | SC Correct answer with no working: B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{B wins and 3 points}) = 0.7\times0.3\times0.3 + 0.3\times0.7\times0.3\) or \(2\times0.063\) \((= 0.126)\) | M2 | ft their diagram; M1 for one correct product or 0.063 |
| \(\frac{P(\text{B wins \& 3 points})}{P(\text{B wins})} = \frac{0.126}{0.216}\) | M1 | Must attempt \(\frac{P(\text{B wins \& 3 points})}{P(\text{B wins})}\); their (c) NOT (d) |
## Question 12(a):
| $0$ | B1 [1] | Allow 0% |
|---|---|---|
---
## Question 12(b):
| Tree diagram with correct structure: $(A)$ branches 0.7, 0.3 leading to $A$/$B$; $(0.7)$ branches 0.7, 0.3; $(0.3)$ branches 0.7, 0.3 leading to $A$/$B$ | B1, B1, B1 [3] | B2: 8 correct branches and probs OR names, no extra branches. B2: 7 correct branches, probs and names, no extra branches. B1: 8 correct branches without probs & names. B1: 6 correct branches, probs and names. Ignore products at ends. Ignore extra branches if no probabilities or $p=0$ |
|---|---|---|
---
## Question 12(c):
| $0.3 \times 0.3 + 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.3$ or $0.09 + 0.063 + 0.063$ | M1 | All correct; M2 ft their diagram |
|---|---|---|
| Two products correct | M1 | ft their diagram |
| $= \frac{27}{125}$ or $0.216$ | A1 [3] | SC Correct answer with no working: B2 |
---
## Question 12(d):
| $0.7 \times 0.3 + 0.3 \times 0.7$ or $0.21 + 0.21$ | M1 | or $1-(0.7^2+0.3^2)$ or $1-(0.49+0.09)$. Condone missing brackets. Or $0.7\times0.3\times0.7 + 0.7\times0.3\times0.3 + 0.3\times0.7\times0.7 + 0.3\times0.7\times0.3$ or $2\times0.147 + 2\times0.063$ |
|---|---|---|
| $= \frac{21}{50}$ or $0.42$ | A1 [2] | SC Correct answer with no working: B1 |
---
## Question 12(e):
| $P(\text{B wins and 3 points}) = 0.7\times0.3\times0.3 + 0.3\times0.7\times0.3$ or $2\times0.063$ $(= 0.126)$ | M2 | ft their diagram; M1 for one correct product or 0.063 |
|---|---|---|
| $\frac{P(\text{B wins \& 3 points})}{P(\text{B wins})} = \frac{0.126}{0.216}$ | M1 | Must attempt $\frac{P(\text{B wins \& 3 points})}{P(\text{B wins})}$; their (c) NOT (d) |12 Anika and Beth are playing a game which consists of several points.
\begin{itemize}
\item The probability that Anika will win any point is 0.7 .
\item The probability that Beth will win any point is 0.3 .
\item The outcome of each point is independent of the outcome of every other point.
\end{itemize}
The first player to win two points wins the game.
\begin{enumerate}[label=(\alph*)]
\item Write down the probability that the game consists of more than three points.
\item Complete the probability tree diagram in the Printed Answer Booklet showing all the possibilities for the game.
\item Determine the probability that Beth wins the game.
\item Determine the probability that the game consists of exactly three points.
\item Given that Beth wins the game, determine the probability that the game consists of exactly three points.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2021 Q12 [13]}}