Question

Bob picks 3 cats. There are 5 striped males, 6 striped females, 8 spotted males, and 4 spotted females.

What is the prob. that Bob has atleast a striped cat given that he has picked exactly one female?

Answer 1

@DanielxAK would I be using combinations here?

Answer 2

I don't think so. You're looking for the probability of something, not how many choices you have. It looks like it'd be similar to what you were doing yesterday.

Actually, wasn't this the problem from yesterday or is it so very similar that I can't immediately see the difference?

Answer 3

This is only different from the other because it asks for the probability of atleast a striped, given Bob picks *Exactly* one female.

Answer 4

Bob really has a thing for cats.

The last one was if he has atleast one female, correct?

Answer 5

haha yes

Answer 6

Hm. I'm not seeing an easy solution. So, maybe someone knows of a better idea on how to approach the problem. But, I'll give you what I have anyway in case it sparks an idea of your own:

Let A be the probability of a stripe, M and N be male, and F be female, then:

P(A|MNF) is what you want. Note: MNF can be in any order, so their probability is equal.

Expanding it, you'd get: P(A|MNF) = [P(MNFA)]/[P(M)P(N|M)P(F|MN)].

Also, note: P(MNFA) = P(MNF)P(A) since stripes and gender are independent of each other.

Yeah, that's about as far as I got. I'm guessing there's an easier way (or I could just be flat out wrong), but this is how I would approach it, even though calculating all those extra probabilities will be tedious.

Answer 7

That makes sense for dependent events in general, but what would be the difference here between the at least part and exactly one, using combinations?

Answer 8

I'll give an example that I think is correct. Again, Probability isn't my strongest subject:

Say you have 5 people and you want to choose exactly 3. Then, you would have 5 choose 3. If you wanted to choose atleast 5 people, you would have 5 choose 3 + 5 choose 4 + 5 choose 5.

Does that make sense?

Answer 9

That makes sense, but I have no idea how to apply that to this conditional problem