Yet another question on my online classes homework that I missed. It doesn't explain how the answer is obtained.
The question:
An urn contains 7 orange balls and 5 blue balls. Two balls are selected at random without replacement. If at least one of them is orange, what is the probability that the other ball is also orange?
The given answer is .375.
So, I'm looking for P(A|B), B is the event that one of them is orange, and A is the event that the other one is also orange. For P(B), I did P(B$^c$) (none of them are orange, which means they are both blue), which would be (5/12) * (4/11), or .1515. P(B) is 1 - P(B$^c$), so P(B) =.8585.
This is where I get stuck. P(A|B) = P(A $\bigcap$ B) / P(B). But I can't figure out how to find P(A $\bigcap$ B).
In the same way as you calculated for the blue ones you get $$P(A \cap B)= \frac{7}{12} \cdot \frac{6}{11}$$ Which gives you the answer after plugging it into the formula you have.