1. **State the problem:** May has 28 candies in 4 colours: orange, blue, green, and purple. Half are orange, blue, or green. Green to purple ratio is 1:2. Blue candies are 1 fewer than green. Find the probability of picking 2 blue candies without replacement. 2. **Set variables:** Let green = $g$, purple = $p$, blue = $b$, orange = $o$. 3. **Write equations:** - Total: $o + b + g + p = 28$ - Half are orange, blue, green: $o + b + g = 14$ - Ratio: $p = 2g$ - Blue fewer than green: $b = g - 1$ 4. **Substitute:** From total: $o + (g - 1) + g + 2g = 28 \Rightarrow o + 4g = 29$ From half: $o + (g - 1) + g = 14 \Rightarrow o + 2g = 15$ 5. **Subtract equations:** $(o + 4g) - (o + 2g) = 29 - 15 \Rightarrow 2g = 14 \Rightarrow g = 7$ 6. **Find others:** $p = 2g = 14$, $b = g - 1 = 6$, $o + 2g = 15 \Rightarrow o = 1$ 7. **Check total:** $1 + 6 + 7 + 14 = 28$ correct. 8. **Calculate probability:** $$P = \frac{\binom{6}{2}}{\binom{28}{2}} = \frac{\frac{6 \times 5}{2}}{\frac{28 \times 27}{2}} = \frac{15}{378} = \frac{5}{126}$$ **Final answer:** $$\boxed{\frac{5}{126}}$$