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Tough Probability Problem [GmatPrep]

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by sarawutc » Mon Jul 22, 2013 3:00 pm
There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15

So from my understanding, the box has 1 green ball, 1 yellow ball, and 8 other balls.
The question is...

It seems that the answer to this question comes from this calculation:
(8c2)(1c1)/(10c3)

Q1. How would I know whether all the 8 balls are the same color ? some of them could be blue and some could be red.?

Q2. If the 8 balls are of the same color. Should the total number of ways to pick 3 balls from this ball be less than 10c3 due to the fact that we have indistinguishable objects?

As an example to illustrate my issue,

If we have ACCC, and we want to choose 2 letters from it
List: AC,CC
Here there are 2 ways.
But 4c2 would mean that there are 6 possible ways

So, back to the original question.
Should the total number of ways to pick 3 balls from this ball be less than 10c3 due to the fact that we have indistinguishable objects?
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by [email protected] » Mon Jul 22, 2013 7:38 pm
Hi sarawutc,

Since you've chosen to think of this probability question in terms of a combination formula question, you might not see the fact that the color of the other 8 balls doesn't matter (because they're not green and they're not yellow). If you were to ignore the combination formula altogether, then here is how you could answer this question...

We're after the probability of getting the 1 green and 2 others that are NOT yellow. ANY of the 3 could be the green one though, so we'd have to account for all the possibilities. So, we could have:

1st = green
2nd = non-yellow
3rd = non-yellow

The probability of THAT happening is 1/10 x 8/9 x 7/8 = 56/720

The green ball could have come second though:

1st = non-yellow, non-green
2nd = green
3rd = non-yellow

The probability of THAT happening is 8/10 x 1/9 x 7/8 = 56/720

Notice how it's the same probability, even though the green ball was chosen second instead of first? What do you think will be the case if the green ball is chosen third?

1st = non-yellow, non-green
2nd = non-yellow, non-green
3rd = green

The probability of THAT happening is 8/10 x 7/9 x 1/8 = 56/720

Now you have ALL THE WAYS to get 1 green and no yellow:

56/720 + 56/720 + 56/720 = 168/720 = 7/30

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by sarawutc » Tue Jul 23, 2013 12:43 am
thank you very much : )
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by GMATGuruNY » Tue Jul 23, 2013 1:05 am
sarawutc wrote:There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15
Here's how we could use combinatorics to solve:

P = (good combinations)/(total possible combinations).

Total combinations:
Number of combinations of 3 can be formed from the 10 balls = 10C3 = (10*9*8)/(3*2*1) = 120.

Good combinations:
In a good combination, 2 NON-YELLOW balls are combined with the green ball.
Of the 9 balls besides the green ball, 8 are NOT yellow.
Number of combinations of 2 that can be formed from the 8 non-yellow balls = 8C2 = (8*7)/(2*1) = 28.
Since there are 28 non-yellow pairs that can be combined with the green ball, there are 28 ways to form a combination of 3 that includes the green ball but not the yellow ball.

Thus:
(good combinations)/(total possible combinations) = 28/120 = 7/30.

The correct answer is B.
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