I assume you are talking about "sampling without replacement"- that is you do NOT put a ball back into the box after noting its color.
1. In order to get "all three ball being red", the first ball must be red. There are a total of 7+ 4+ 9= 20 ball, 7 of them red so the probability of that is 7/20.
Once that has happened, there are 19 balls left, 6 of them red. The probability that the second ball you pick will be red is 6/19. The probability that the first two balls are both red is the product: (7/20)(6/19).
If you have done that, then there are 18 balls left, 5 of them red: the probability that the next ball is also red is 5/18. The probability of picking 3 red balls is
(7/20)(6/19)(5/18) which can be reduced (since 5/20= 1/4 and 6/18= 1/3) to
(7/4)(1/19)(1/3)= 7/228 or about 0.0307.
"2 white and one red" is just a little harder. One way for that to happen is to get "white, white, red" in that order. Since there are 4 white balls, the probability that the first ball will be white is 4/20= 1/5. If the first ball is white, then there are 19 balls left and 3 of them are white. The probability that the second ball is also white is 3/19. Assuming that the first two balls were white, there are now 18 balls left but 7 of them are still red. The probability that the third ball will be red is 7/18. The probability of gettin "white,white,red", in that order, is (1/5)(3/19)(7/18).
But drawing "white, red, white" would also give "2 white and 1 red". As before the probability that the first ball is white is 4/20= 1/5. Assuming the first ball is white then on the second draw there are 19 balls left and 7 of them are red. The probability that the second ball is red is 7/19. Finally, if the draws have happened that way there are 18 balls left, of which 3 are still white. The probability that the third ball is white is 3/18= 1/6. The probability of getting "white, red, white" in that order is (1/5)(7/19)(3/18). Notice that the denominators are exactly the same as in the previous case! Of course: that was always the total number of balls and it didn't matter what color was taken out. Even more, the numerators are the same- just their order is changed. That's because we are taking out the same balls- just changing the order. But multiplication is "commutative"- order doesn't matter. The point is that the probability of getting these balls in any specific order is the same: (1/5)(3/19)(7/18). To find the probability of getting "white, white, red" is any order, we just have to recognize that there are 3 "orders": (white, white, red), (white, red, white), and (red, white, white) so the probability of getting 2 white and 1 red ball, in any order is 3(1/5)(3/19)(7/18)= (1/5)(1/19)(7/2)= 7/190 or about 0.0368.
