**5(a)(i)** Require $R = 3$ and $G = 4$ so probability is $\frac{3}{4} \times \frac{1}{3}$ | M1 | AO 2.1

$= \frac{1}{4}$ or **0.25** | A1 | AO 1.1b

**5(a)(ii)** $[R \text{ must be } 2 \text{ and } G = 1 \text{ so } \frac{1}{4} \times \frac{2}{3} = ] \frac{1}{6}$ | A1 | AO 1.1b

**Total: 3 marks**

**5(b)** $P(X = 50) = 0.25$ must mean $R = 3$ and $G = 4$ so $3m + 4n = 50$ | M1 | AO 3.1a
| A1 | AO 1.1b

$P(X = 20) = \frac{1}{6} \Rightarrow R = 2, G = 1$ so $2m + n = 20$ | A1 | AO 2.1

Solving: $3m + 4(20 - 2m) = 50$ (o.e.) | M1 | AO 1.1b

**$m = 6$ and $n = 8$** | A1 | AO 3.2a

**Total: 5 marks**

**Grand Total: 8 marks**

**Notes:**

**(a)(i)** M1 for sight of $\frac{3}{4} \times \frac{1}{3}$ or $\frac{1}{4} \times \frac{2}{3}$ as a single product BUT allow e.g. $\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}$ to score M1. However if the products are later added e.g. $\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}$ it is M0. May be implied by one correct answer to (i) or (ii).

A1 for $\frac{1}{4}$ or 0.25 or exact equivalent (allow 25%)

**(a)(ii)** A1 for $\frac{1}{6}$ or exact equivalent

**(b)** For the **1st 4 marks** condone incorrect labelling e.g. $R$ for $m$ or $G$ for $n$ if intention is clear.

1st M1 for identifying either set of cases $(R = 2, G = 1, X = 20)$ or $(R = 3, G = 4, X = 50)$. Allow 1st M1 for $P(X = 20) = \frac{1}{4} \times \frac{3}{4}$ or $P(X = 50) = \frac{3}{4} \times \frac{1}{4}$ NOT just $P(X = 20) = \frac{1}{6}$ etc.

or $\frac{1}{4}m + \frac{3}{4}n = 20$ or $\frac{3}{4}m + \frac{1}{4}n = 50$ and might score 2nd M1 (answer is $m = 64, n = 6$)

or $\frac{1}{4}m + \frac{3}{4}n = \frac{1}{4}$ or $\frac{3}{4}m + \frac{1}{4}n = \frac{1}{4}$ and might score 2nd M1 (answer is $m = \frac{5}{15}, n = \frac{-5}{20}$)

or $2m + n = \frac{1}{4}$ or $3m + 4n = \frac{1}{4}$ and might score 2nd M1 (answer is $m = \frac{5}{12}, n = 0$)

or $2m + n = 50$ **and** $3m + 4n = 20$ and might score 2nd M1 (answer is $m = 36, n = -22$)

1st A1 for one correct equation

2nd A1 for **both** correct equations and no incorrect equations, unless they attempt to solve the correct 2 equations only

2nd M1 for attempt to solve their linear equations in $m$ and $n$ (reduce to an equation in one variable, condone one sign error). May be implied by $m = 6$ and $n = 8$.

**Calc** If they use one of the 4 sets of equations for 1st M1 and use a calculator to write down the answer, we will allow this mark for sight of the correct answers to those equations as given above.

3rd A1 $m = 6$ and $n = 8$ only (no incorrect labelling here). Correct answer by trial can score 5/5 if no incorrect working seen.