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?

## Difficult probability

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- joannabanana
- Master | Next Rank: 500 Posts
**Posts:**118**Joined:**05 Sep 2010**Location:**Canada**Thanked**: 5 times**GMAT Score:**730

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

- GMATGuruNY
- GMAT Instructor
**Posts:**15532**Joined:**25 May 2010**Location:**New York, NY**Thanked**: 13060 times**Followed by:**1897 members**GMAT Score:**790

You determined P(RRBG) = 1/35.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?

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.

Mitch Hunt

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