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Probability Approaches

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by tnkippen » Tue Nov 13, 2012 7:38 am
I have been trying to use combinations more to solve probability questions, but I am finding a discrepancy between the combinations approach and the traditional approach on the problem below.

"There are 6 balls: A, B, C, D, E, F. What is the probability that a package containing two different types of balls contains E?"

Using combinations, I calculated: (5C5)(5C1)/(6C2) = 5/15 = 1/3

But using P(A or B) = P(A) + P(B) - P(A and B), I got:
P(A) = 1/6, P(B) = 1/5, P(A and B) = 0, so:
(1/6) + (1/5) = 11/30

What am I doing wrong above?
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by Brent@GMATPrepNow » Tue Nov 13, 2012 8:08 am
tnkippen wrote:I have been trying to use combinations more to solve probability questions, but I am finding a discrepancy between the combinations approach and the traditional approach on the problem below.

"There are 6 balls: A, B, C, D, E, F. What is the probability that a package containing two different types of balls contains E?"

Using combinations, I calculated: (5C5)(5C1)/(6C2) = 5/15 = 1/3

But using P(A or B) = P(A) + P(B) - P(A and B), I got:
P(A) = 1/6, P(B) = 1/5, P(A and B) = 0, so:
(1/6) + (1/5) = 11/30

What am I doing wrong above?
The problem is highlighted above (in green)

P(E is 1 of the 2 balls) = P(E selected first OR E selected second)
= P(E selected first) + P(E selected second) - P(E selected first and second)
= P(E selected first) + P(E selected second) - 0 (since E cannot be selected 1st and 2nd)

P(E selected first) = P(E selected 1st AND some other letter selected 2nd)
= P(E selected 1st) x P(some other letter selected 2nd)
= (1/6)(5/5)
= 1/6

P(E selected second) = P(some other letter selected 1st AND E selected 2nd)
= P(some other letter selected 1st) x P(E selected 2nd)
= (5/6)(1/5)
= 1/6

So, P(E is 1 of the 2 balls) = 1/6 + 1/6 = 1/3

Cheers,
Brent
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by tnkippen » Tue Nov 13, 2012 8:15 pm
Thanks again, Brent. With the combinations approach, why don't I need to multiply (5C5)(5C1)/(6C2) by 2 (to represent the chance that the E could occur first or second)?
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by sana.noor » Sun Oct 27, 2013 11:39 pm
we can select 2 balls from 6 balls as 6C2 = 15 ways
now withdraw E, and rest we are left with 5 balls. drawing one ball from rest of the five balls 5C1 = 5 ways. now E is already selected. 5/15 = 1/3 ways we can combine one ball out of 5 different balls with E.
Isnt it too easy?
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by GMATGuruNY » Mon Oct 28, 2013 2:52 am
There are 6 balls: A, B, C, D, E, F. What is the probability that a package containing two different types of balls contains E?
Alternate approach:

P(E is selected) = 1 - P(E is NOT selected).

P(E is not selected):
P(1st ball is not E) = 5/6. (Of the 6 balls, 5 are not E.)
P(2nd ball is not E) = 4/5. (Of the 5 remaining balls, 4 are not E.)
Since both events must happen for E not to be selected, we MULTIPLY the probabilities:
5/6 * 4/5 = 2/3.

Thus:
P(E is selected) = 1 - 2/3 = 1/3.
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