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## Comment on

Same Color## Can we solve the following

A bag holds 4 red marbles, 5 blue marbles, and 2 green marbles. If 5 marbles are selected one after another without replacement, what is the probability of drawing 2 red marbles, 2 blue marbles, and 1 green marble?

## No, you'd have to use a

No, you'd have to use a different approach, since the required outcomes for each draw aren't identical. In the question above, we must draw a red ball each time.

In your question, there are several possible orderings (e.g., red - red - blue - blue - green, OR blue - red - green - blue - red, OR ...

In my opinion, such a question would be out of scope (too difficult) for the GMAT.

## Now I get the point.

Thanks a lot.

## what if we use the

## That approach works also.

That approach works also.

## I followed this approach and

P (same color)= P(1st draw any color)*P(2nd draw same color)

Case 1: 1st draw was green. P(any color)* P(same color) = 1*3/8= 3/8

Case 2: 1st draw was red. P (any color)* P(same color)= 1*2/8= 2/8

Case 3: 1st draw was yellow. P (any color)* P(same color)= 1*1/8= 1/8

Total= case1+case2+case3

3/8 + 2/8 + 1/8 = 6/8 = 3/4

What is wrong with this approach?

## Your cases don't match your

Your cases don't match your probabilities.,

For example, "Case 1: 1st draw was green. P(any color)* P(same color) = 1*3/8= 3/8"

If the 1st draw is green, then the first draw can't be "any color." The first draw must be green.

Here's what you need to do....

Case 1: 1st draw is green: P(1st is green) x P(2nd is green) = (4/9) x (3/8) = 12/72

Case 2: 1st draw is red. P(1st is red) x P(2nd is red) = (3/9) x (2/8) = 6/72

Case 3: 1st draw is yellow. P(1st is yellow) x P(2nd is yellow) = (2/9) x (1/8) = 2/72

etc..

## Is this approach valid?

(4C2 + 3C2 + 2C2)/9C2 = 10/36 = 5/18

## Perfect!!

Perfect!!

## Add a comment