There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue?
I solved it this way:
3/7 * 2/6 * 2/5 * 2/4 = 1/35
thinking that the solution is: p(r) * p(r) * p(g) * p(b)
and the numbers decrease after each selection because the balls are picked without replacement.
The answer however, is 12/35. Can someone please explain why?
I solved it this way:
3/7 * 2/6 * 2/5 * 2/4 = 1/35
thinking that the solution is: p(r) * p(r) * p(g) * p(b)
and the numbers decrease after each selection because the balls are picked without replacement.
The answer however, is 12/35. Can someone please explain why?
