There are $\binom{6}{2}=15$ ways to choose Any $2$ crayons. There are $3$ ways to choose $2$ red crayons (call them $r_1, r_2, r_3$, then $(r_1,r_2)$, $(r_2,r_3)$ and $(r_1,r_3)$ are the ways). So the answer is $\frac{3}{15}=\frac{1}{5}$.