# Math Help - Help with combinatorics problem

1. ## Help with combinatorics problem

This is probably a simple problem for most you guys but I'm only in an elementary probability and statistics class here at college. My professor gave us this problem as practice for our exam. I would really like to solve it. Here it is:

A. A bag as 12 balls, 6 are orange and 4 are blue. 3 unknown balls are removed form the bag. If you remove a 4th ball, what is the probability the 4th ball is blue?

B. With the same bag, find the probability all the removed balls are orange if it is known that at least one of the balls is orange.

For part A, I think I know how to partially set it up. You want to find all of the probabilities of the 3 unknown balls( all blue, 2 blue and 1 orange, 2 orange and 1 blue, etc.) From there I'm stuck and I know there is a more "elegant" way to set it up.

I'm not sure at all about B.

2. Originally Posted by belle05
This is probably a simple problem for most you guys but I'm only in an elementary probability and statistics class here at college. My professor gave us this problem as practice for our exam. I would really like to solve it. Here it is:

A. A bag as 12 balls, 6 are orange and 4 are blue. 3 unknown balls are removed form the bag. What is the probability the 4th ball is blue?

B. With the same bag, find the probability all the removed balls are orange if it is known that at least one of the balls is orange.

For part A, I think I know how to partially set it up. You want to find all of the probabilities of the 3 unknown balls( all blue, 2 blue and 1 orange, 2 orange and 1 blue, etc.) From there I'm stuck and I know there is a more "elegant" way to set it up.
It is exactly the same as the probability of the first ball drawn being blue.

CB

3. Ok, I'm more concerned with setting it up than I am with the actually answer. Any way you could help me out there?

4. Originally Posted by belle05
Ok, I'm more concerned with setting it up than I am with the actually answer. Any way you could help me out there?
Initially, a third of the balls are blue.

If 3 balls are removed, that's one-quarter of the 12 removed,
hence the expected number of blue balls remaining is 3,
since a quarter of 4 is 1.

Hence, with 9 balls remaining, the fraction of these that are blue is still expected to be a third.

5. Thanks!!!!

How about part B. Any pointers?

6. Originally Posted by belle05
Thanks!!!!

How about part B. Any pointers?
You can use conditional probability for the second one if you wish.

First find the probability of at least one of the 3 being orange.

This is 1-P(none are orange).

Then use the conditional probability formula.