Are you referring to the balls question --
the answer for P(X = 1) would be 5c2 x 4c1/9c3 ( as we are drawing out 2 white colored balls from 5 balls and 1 green colored ball out of 4 balls, and the total ways in which the three balls can be drwan is 9c3)
In the other ways, this can also be thought of as first drawing 1 white ball, then in the second attempt drawing another white ball and then in the third attempt drawing a green ball i.e 5c1 x 4c1 x 4c1, however, as you would have noticed, this would mean that there are 3 different permutations possible -- 1 green ball can either be drawn in the first draw, second draw or the third draw. And hence, you would have to multiply this entire thing with 3 to account for that.
So, in that case the P(X= 1) = 3 x 5c1 x 4c1 x 4c1/9c1 x 8c1 x 7c1 = 10/21 (same result).