| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Transfer between containers |
| Difficulty | Standard +0.8 This problem requires constructing a tree diagram with multiple branches, working backwards from a constraint (equal numbers of each color) to determine n, then calculating compound probabilities. It demands systematic case analysis, algebraic manipulation with an unknown parameter, and careful probability calculations across multiple pathways—significantly more complex than routine tree diagram exercises. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Must end up with 3 of each colour or 4 of each colour | M1 | 1st M1 for an overall strategy realising there are 2 options. Award when evidence of both cases (3 of each colour or 4 of each colour) seen. |
| \(n = 2\) requires 1st red and 2nd green or red from A and green from B | M1 | 2nd M1 for \(n = 2\) and attempt at 1st red and 2nd green. May be implied by e.g. \(\frac{4}{9} \times \frac{1}{9}\) |
| \(P(\text{1st red and 2nd green}) = \frac{4}{9} \times \frac{1}{10} = \frac{4}{90}\) or \(\frac{2}{45} \quad p = \dfrac{2}{45}\) | A1 | 1st A1 for \(p = \dfrac{2}{45}\) or exact equivalent |
| \(n = 5\) requires 1st green and 2nd yellow or green from A and yellow from B | M1 | 3rd M1 for \(n = 5\) and attempt at 1st green and 2nd yellow. May be implied by e.g. \(\frac{5}{12} \times \frac{3}{9}\) |
| \(P(\text{1st green and 2nd yellow}) = \frac{5}{12} \times \frac{3}{10} = \frac{15}{120}\) or \(\frac{1}{8} \quad p = \dfrac{1}{8}\) | A1 | 2nd A1 for \(p = \dfrac{1}{8}\) or exact equivalent. NB If both correct values of \(p\) are found and then added (get \(\frac{61}{360}\)), deduct final A1 only (i.e. 4/5) |
# Question 5:
## Full solution
Must end up with 3 of each colour or 4 of each colour | M1 | 1st M1 for an overall strategy realising there are 2 options. Award when evidence of both cases (3 of each colour or 4 of each colour) seen.
$n = 2$ requires 1st red and 2nd green or red from **A** and green from **B** | M1 | 2nd M1 for $n = 2$ and attempt at 1st red and 2nd green. May be implied by e.g. $\frac{4}{9} \times \frac{1}{9}$
$P(\text{1st red and 2nd green}) = \frac{4}{9} \times \frac{1}{10} = \frac{4}{90}$ or $\frac{2}{45} \quad p = \dfrac{2}{45}$ | A1 | 1st A1 for $p = \dfrac{2}{45}$ or exact equivalent
$n = 5$ requires 1st green and 2nd yellow or green from **A** and yellow from **B** | M1 | 3rd M1 for $n = 5$ and attempt at 1st green and 2nd yellow. May be implied by e.g. $\frac{5}{12} \times \frac{3}{9}$
$P(\text{1st green and 2nd yellow}) = \frac{5}{12} \times \frac{3}{10} = \frac{15}{120}$ or $\frac{1}{8} \quad p = \dfrac{1}{8}$ | A1 | 2nd A1 for $p = \dfrac{1}{8}$ or exact equivalent. **NB** If both correct values of $p$ are found and then added (get $\frac{61}{360}$), deduct final A1 only (i.e. 4/5)\begin{enumerate}
\item Two bags, $\mathbf { A }$ and $\mathbf { B }$, each contain balls which are either red or yellow or green.
\end{enumerate}
Bag A contains 4 red, 3 yellow and $n$ green balls.\\
Bag $\mathbf { B }$ contains 5 red, 3 yellow and 1 green ball.\\
A ball is selected at random from bag $\mathbf { A }$ and placed into bag $\mathbf { B }$.\\
A ball is then selected at random from bag $\mathbf { B }$ and placed into bag $\mathbf { A }$.\\
The probability that bag $\mathbf { A }$ now contains an equal number of red, yellow and green balls is $p$.
Given that $p > 0$, find the possible values of $n$ and $p$.
\hfill \mbox{\textit{Edexcel AS Paper 2 2021 Q5 [5]}}