Thread: Conditional probability

Suppose you have an urn with 2 yellow balls, 3 red balls, 6 white balls, and 12 blue balls. You randomly draw 4 balls from the urn without replacement. Given that at least one of the balls is yellow, what is the probability that all 4 balls will be different colors?

I know that you have to use Bayes' rule, with P(B|A) = 1 because if you have 4 colors, one of them must be yellow. But I'm not sure about the different probabilities.

I think P(A) might be 2 choose 1 + 3 choose 1 + 6 choose 1 + 12 choose 1 all divided by 23 choose 4, but I"m not sure. And I don't know how to get P(>= 1 yellow).Any one have any idea?

Thanks!

Hello, horan!

Suppose you have an urn with 2 yellow balls, 3 red balls, 6 white balls, and 12 blue balls.
You randomly draw 4 balls from the urn without replacement.
Given that at least one of the balls is yellow,
what is the probability that all 4 balls will be different colors?

Let = "4 different colors" . . .and = "at least one yellow"

According to Bayes' Theorm, we want . .[1]

There are: . possible drawings.


Numerator

As you pointed out:

The number of ways to get 4 colors is: . ways.
Hence: .


Denominator

The opposite of "at least 1 yellow" is "no yellow."
There are: . ways to get no yellows.
So there are: . ways to get at least one yellow.

Hence: .


Substitute into [1]: .