1. **State the problem:** May has 28 candies in 4 different colours: orange, blue, green, and purple. Half of the candies are orange, blue, or green. The ratio of green to purple candies is 1:2. There is 1 fewer blue candy than green candy. We need to find the probability that if she picks 2 candies randomly without replacement, both are blue. 2. **Define variables:** Let the number of green candies be $g$, purple candies be $p$, blue candies be $b$, and orange candies be $o$. 3. **Use given information:** - Total candies: $$o + b + g + p = 28$$ - Half are orange, blue, or green: $$o + b + g = \frac{28}{2} = 14$$ - Ratio of green to purple: $$g : p = 1 : 2 \implies p = 2g$$ - Blue candies are 1 fewer than green: $$b = g - 1$$ 4. **Express orange candies:** From total candies, $$o + b + g + p = 28$$ Substitute $p = 2g$ and $b = g - 1$: $$o + (g - 1) + g + 2g = 28$$ Simplify: $$o + 4g - 1 = 28$$ $$o + 4g = 29$$ 5. **Use half candies condition:** $$o + b + g = 14$$ Substitute $b = g - 1$: $$o + (g - 1) + g = 14$$ $$o + 2g - 1 = 14$$ $$o + 2g = 15$$ 6. **Solve system of equations:** From step 4 and 5: $$\begin{cases} o + 4g = 29 \\ o + 2g = 15 \end{cases}$$ Subtract second from first: $$ (o + 4g) - (o + 2g) = 29 - 15 $$ $$ 2g = 14 $$ $$ g = 7 $$ 7. **Find other values:** $$ p = 2g = 2 \times 7 = 14 $$ $$ b = g - 1 = 7 - 1 = 6 $$ $$ o + 2g = 15 \implies o + 14 = 15 \implies o = 1 $$ 8. **Check total:** $$ o + b + g + p = 1 + 6 + 7 + 14 = 28 $$ correct. 9. **Calculate probability:** Probability of picking 2 blue candies without replacement: $$ P = \frac{\binom{6}{2}}{\binom{28}{2}} = \frac{\frac{6 \times 5}{2}}{\frac{28 \times 27}{2}} = \frac{15}{378} = \frac{5}{126} $$ **Final answer:** $$\boxed{\frac{5}{126}}$$