Sample space ={R,R,R,R,R,W,W,W,W,W,W,W} = (5 R + 7 W = 12)
1st selection: P(one red) = 5/12 = 0.4167. Now the remaining cubes after the 1st selection : 4 Red and 7 Whites = 11 cubes), so
2nd .selection P(one red) = 4/11 = 0.3667
This is a conditional probability and the condition is ; GET 2 REDS.
Hence P(getting one red AND getting also a red) = 5/12 x 4/11 =20/132=5/33 or 0.15