Difficult probability

Solved it using combinations:

Looks like this: [(3C2)(2C1)(2C1)] / (7C4) = 12/35

Explanation of Terms:
1st term- # ways to arrange 3 Red Marbles in 2 picks: 3C2 = 3!/(1!)(2!) = 3

2nd term - # ways to arrange 2 Blue marbles in 1 pick: 2C1 = 2!/(1!)(1!) = 2

3rd term- # ways to arrange 2 Green marbles in 1 pick: 2C1 = 2!/(1!)(1!) = 2

Denominator- How many ways to arrange 7 marbles in 4 picks: 7C4= 7!/(4!)(3!) = 35
So 35 ways to arrange 4 marbles out of 7.

(3)(2)(2)/35 = 12/35


Your method assumed the marbles were picked in the order of RED, RED, GREEN, BLUE , but there are actually 12 ways we could pick these:

RRGB RGBR RBGR RRBG RBRG RGRB
GRRB GBRR GRBR BGRR BRGR BRRG

joannabanana wrote: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?

You determined P(RRBG) = 1/35.

But RRBG is only 1 way to get a good outcome. To account for all the ways we could get 2 R's, 1 B and 1 G, we need to multiply by the number of ways to arrange RRBG = 4!/2! = 12.

12*1/35 = 12/35.