Approach #1: Probability rules
P(different colors) = P(red 1st AND blue 2nd OR blue 1st AND red 2nd

= P(red 1st AND blue 2nd) + P(blue 1st AND red 2nd)

= [P(red 1st) x P(blue 2nd)] + [P(blue 1st) x P(red 2nd)]

= [$\frac{r}{r+b}$ x $\frac{b}{r+b-1}$] + [$\frac{b}{r+b}$ x $\frac{r}{r+b-1}$]

= $\frac{br}{(b+r)(b+r−1)}$ + $\frac{br}{(b+r)(b+r−1)}$

= $\frac{2br}{(b+r)(b+r−1)}$

Answer: D

Approach #2: Testing values
Notice that, if $r=1$ and $b=1$, then we are guaranteed to have two different colors.
In other words, $r=1$ and $b=1$, then P(different colors) = $1$

So now we'll plug $r=1$ and $b=1$ into each answer choice to see which one equals $1$....

A) $\frac{(1)(1)}{(1+1)(1+1-1)}=\frac{1}{2}$. NO GOOD.

B) $\frac{2(1+1)}{(1+1)(1+1-1)}=2$. NO GOOD.

C) $\frac{(1)(1)}{(1+1)}=\frac{1}{2}$. NO GOOD.

D) $\frac{2(1)(1)}{(1+1)(1+1-1)}=1$. BINGO!!!

E) $\frac{2(1)(1)}{(1+1)(1+1)}=\frac{1}{2}$. NO GOOD.

Answer: D