| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram question with clearly defined probabilities (1/10, 1/5, 1/3 for red) and standard conditional probability calculations. Part (d) requires Bayes' theorem but in a routine context. The stopping condition adds minimal complexity as the structure is explicitly described. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct tree diagram shape (3 pairs) with at least one label on at least two pairs G(reen) and R(ed) | B1 | Allow G and G' or R and R' as labels; condone 'extra' pairs if labelled with probability of 0 |
| All 6 correct probabilities on correct branches with at least one label on each pair | dB1 | Dependent on previous B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{9}{10} \times \frac{4}{5} \times \frac{2}{3}\) | M1 | Multiplication of 3 correct probabilities (allow ft from their tree diagram) |
| \(= \frac{12}{25}\) \((= 0.48)\) | A1 | \(\frac{12}{25}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{9}{10} \times \frac{1}{5} + \frac{9}{10} \times \frac{4}{5} \times \frac{1}{3}\) or \(1 - \left(\frac{1}{10} + \frac{9}{10} \times \frac{4}{5} \times \frac{2}{3}\right)\) | M1 | Either addition of only two correct products (product of two probs + product of three probs) which may ft from tree diagram, or for \(1 - (\frac{1}{10} + {'}(b){'}\)) |
| \(= \frac{21}{50}\) \((= 0.42)\) | A1 | \(\frac{21}{50}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([P(\text{Red from } B \mid \text{Red selected})] = \dfrac{\frac{9}{10} \times \frac{1}{5}}{\frac{1}{10} + \frac{9}{10} \times \frac{1}{5} + \frac{9}{10} \times \frac{4}{5} \times \frac{1}{3}}\) | M1 | Correct ratio of probabilities or correct ft ratio e.g. \(\dfrac{\frac{9}{10} \times \frac{1}{5}}{1 - {'}(b){'}}\\) or \(\dfrac{\frac{9}{10} \times \frac{1}{5}}{\frac{1}{10} + {'}(c){'}}\\) with numerator \(<\) denominator |
| \(= \dfrac{9}{26}\) | A1 | Allow awrt \(0.346\) |
## Question 1:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct tree diagram shape (3 pairs) with at least one label on at least two pairs G(reen) and R(ed) | B1 | Allow G and G' **or** R and R' as labels; condone 'extra' pairs if labelled with probability of 0 |
| All 6 correct probabilities on correct branches with at least one label on **each** pair | dB1 | Dependent on previous B1 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{9}{10} \times \frac{4}{5} \times \frac{2}{3}$ | M1 | Multiplication of 3 correct probabilities (allow ft from their tree diagram) |
| $= \frac{12}{25}$ $(= 0.48)$ | A1 | $\frac{12}{25}$ oe |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{9}{10} \times \frac{1}{5} + \frac{9}{10} \times \frac{4}{5} \times \frac{1}{3}$ **or** $1 - \left(\frac{1}{10} + \frac{9}{10} \times \frac{4}{5} \times \frac{2}{3}\right)$ | M1 | Either addition of only two correct products (product of two probs + product of three probs) which may ft from tree diagram, or for $1 - (\frac{1}{10} + {'}(b){'}$) |
| $= \frac{21}{50}$ $(= 0.42)$ | A1 | $\frac{21}{50}$ oe |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[P(\text{Red from } B \mid \text{Red selected})] = \dfrac{\frac{9}{10} \times \frac{1}{5}}{\frac{1}{10} + \frac{9}{10} \times \frac{1}{5} + \frac{9}{10} \times \frac{4}{5} \times \frac{1}{3}}$ | M1 | Correct ratio of probabilities **or** correct ft ratio e.g. $\dfrac{\frac{9}{10} \times \frac{1}{5}}{1 - {'}(b){'}}\$ or $\dfrac{\frac{9}{10} \times \frac{1}{5}}{\frac{1}{10} + {'}(c){'}}\$ with numerator $<$ denominator |
| $= \dfrac{9}{26}$ | A1 | Allow awrt $0.346$ |
> **Note:** Allow decimals or percentages throughout this question. Total: **(8 marks)**\begin{enumerate}
\item Three bags, $A , B$ and $C$, each contain 1 red marble and some green marbles.
\end{enumerate}
Bag $A$ contains 1 red marble and 9 green marbles only\\
Bag $B$ contains 1 red marble and 4 green marbles only\\
Bag $C$ contains 1 red marble and 2 green marbles only\\
Sasha selects at random one marble from bag $A$.\\
If he selects a red marble, he stops selecting.\\
If the marble is green, he continues by selecting at random one marble from bag $B$.\\
If he selects a red marble, he stops selecting.\\
If the marble is green, he continues by selecting at random one marble from bag $C$.\\
(a) Draw a tree diagram to represent this information.\\
(b) Find the probability that Sasha selects 3 green marbles.\\
(c) Find the probability that Sasha selects at least 1 marble of each colour.\\
(d) Given that Sasha selects a red marble, find the probability that he selects it from bag $B$.
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q1 [8]}}