Question: Same Color

Comment on Same Color

Can we solve the following question using the same approach?

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?
gmat-admin's picture

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 probability basic formula combined with counting/listing to find the numerator=(6+3+1)=10 and the denominator=36, and the p=10/36=5/18. I think this would be an easier solution.
gmat-admin's picture

That approach works also.

I followed this approach and I got a wrong answer:

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?
gmat-admin's picture

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
gmat-admin's picture

Perfect!!

If I were to list the possibilities for the numerator, why isn't it just 3 (GG, RR, YY)? what am i missing here?
gmat-admin's picture

The main issue is that those three outcomes (GG, RR, YY) are not all equally likely.

To illustrate this, let's say there are 1,000,000 green balls and 2 red balls.
As you can imagine, the likelihood of getting 2 green balls is much much much greater than the likelihood of selecting 2 red balls.

Don't forget that, if we're going to use counting techniques to solve a probability question, each outcome must be equally likely.
For more see: https://www.gmatprepnow.com/module/gmat-probability/video/742

Here's another example: A barrel contains 999 red marbles and 1 blue marble. If one marble is selected at random from the barrel, what is the probability that the selected marble is blue?
If we list the possible outcomes the way you did we get two possibilities:
- Red
- Blue
Since 1 of the 2 possible outcomes is blue, one might incorrectly conclude that P(the selected ball is blue) = 1/2

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