# Thread: Math stats and probability help

1. ## Math stats and probability help

Urn I contains 13 red balls and 11 blue balls. Urn II contains 14 red balls and 7 blue balls. Four balls are drawn at random from urn I and placed in urn II. Then five balls are
drawn from urn II. What is the probability that two of the balls drawn from urn I were blue and the
other two were red, given that all the balls drawn from urn II were the same color?

2. ## Re: Math stats and probability help

You can give us your answer and be told whether or not it is correct.

You can show us your completed work and be told whether it is correct and, if not correct, exactly where and why.

You can show us your uncompleted work and be told what is the next step.

You can ask for a hint to get started if you tell us what has you confused.

In this case, you would not have been given this problem without knowing something about probability so we cannot even tell what seems difficult to you about the problem. Are you familiar with conditional probabilities.

3. ## Re: Math stats and probability help

Originally Posted by ballerninja29
Urn I contains 13 red balls and 11 blue balls. Urn II contains 14 red balls and 7 blue balls. Four balls are drawn at random from urn I and placed in urn II. Then five balls are
drawn from urn II. What is the probability that two of the balls drawn from urn I were blue and the
other two were red, given that all the balls drawn from urn II were the same color?
Welcome to this forum. As a new member please know that this is not a homework service, we are here to help you learn.
So you must work with us.

In this question we must step back and see that it is counting question. Next we must decide what model helps us here. In this I would say it is similar to selecting men and women, even though the balls are identical except for colour.
The first draw can consist of $R_1=0\text{ to }4$ red balls which means there are corresponding $B_1=4-R_1$ blue balls. Now, each of those 'draws' has a different probability: Ex. $\mathscr{P}(R_1=2)=\dfrac{\binom{13}{2}\cdot \binom{11}{2}}{\binom{24}{4}}$

To finish you must work with a changed population in urn II. If $R_1=2$ then how many red & blue balls are now in urn II?

Show us some work.