A bag contains 10 red, 7 blue, and 3 green disks.
The probability of choosing a red disk at random is 10/20= 1/2. The probability that all three pick a red disk is (1/2)^3= 1/8..
Reggie takes 3 disk out of the bag, not returning them to the bag.
What is the probability that all three disks are of different color.
The probability the first disk is red is 10/20= 1/2. There are then 9 red, 7 blue, and 3 green disk. The probability the second disk is blue is 7/19. There are the 9 red, 6 blue, and 3 green. The probability the third disk is green is 3/18= 1/6. The probability of "red, blue, green" in that order, is (1/2)(7/19)(1/6)= 7/228.
The probability the first disk is red is 10/20= 1/2. There are then 9 red, 7 blue, and 3 green disk. The probability the second disk is green is 3/19. There are the 9 red, 7 blue, and 2 green. The probability the third disk is blue is 7/18. The probability of "red, blue, green" in that order, is (1/2)(3/19)(7/19)= 7/228.
The probability the first disk is blue is 7/20. There are then 10 red, 6 blue, and 3 green disk. The probability the second disk is red is 10/19. There are the 9 red, 6 blue, and 3 green. The probability the third disk is green is 3/18= 1/6. The probability of "red, blue, green" in that order, is (7/20)(10/19)(1/6)= 7/228.
Get the point? There are 3!= 6 different orders of "red, green, and blue" in any order so the probability of "all three different colors" is 6(7/228)= 7/38.