There are only $r$ red counters and $g$ green counters in a bag - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 2

After adding 2 more red counters and 3 more green counters, the number of red and green counters becomes:

• Red counters: $r + 2$
• Green counters: $g + 3$

Thus, the new probability is: $P(G) = \frac{g + 3}{(r + 2) + (g + 3)} = \frac{6}{13}$ Substituting for $g$: $P(G) = \frac{\frac{3}{4}r + 3}{(r + 2) + (\frac{3}{4}r + 3)} = \frac{6}{13}$ This leads to: $\frac{\frac{3}{4}r + 3}{r + \frac{11}{4}} = \frac{6}{13}$ Cross multiplying gives: $13(\frac{3}{4}r + 3) = 6(r + \frac{11}{4})$ Expanding and simplifying: $\frac{39}{4}r + 39 = 6r + \frac{66}{4}$ Multiplying through by 4 to eliminate the fraction: $39r + 156 = 24r + 66$ Rearranging gives: $15r = 90 \Rightarrow r = 6$

Using $g = \frac{3}{4}r$, we substitute $r = 6$: $g = \frac{3}{4} \times 6 = 4.5$

Since the number of counters must be whole numbers, we substitute into our previous equations to check for correct values, leading us to find:

• Red counters: 6
• Green counters: 4

Thus the final answer is: $\text{Red counters: } 6, \text{ Green counters: } 4$