# Math Help - More conditional prob

1. ## More conditional prob

Suppose there are two bags A and B and a set of colored balls (red and green).

Partially color-blind people are told to place all the red balls in A and the green balls in B.

Their ability to distinguish the balls is 80% reliable.

15% of the balls are red, the rest green. (85%)

If we randomly choose a ball from bag A, what is the probability of it being red? (or green)?

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My work:

We know that P(R) = .15, P(G) = .85.

I think we need something like P(R|A) = ?

Not sure how to proceed

EDIT: Ah! I think I see. I think that P(A | R) = .8 and P(B|R)=.2, etc... is that right?

EDIT 2: Hmm.. no, still not sure what P(A) should be though...

2. Originally Posted by scorpion007
Suppose there are two bags A and B and a set of colored balls (red and green).

Partially color-blind people are told to place all the red balls in A and the green balls in B.

Their ability to distinguish the balls is 80% reliable.

15% of the balls are red, the rest green. (85%)

If we randomly choose a ball from bag A, what is the probability of it being red? (or green)?

-------------

My work:

We know that P(R) = .15, P(G) = .85.

I think we need something like P(R|A) = ?

Not sure how to proceed

EDIT: Ah! I think I see. I think that P(A | R) = .8 and P(B|R)=.2, etc... is that right?

EDIT 2: Hmm.. no, still not sure what P(A) should be though...
Love these questions:
My thoughts:
Let's assume there are 100 balls altogether - 15 red and 85 green. We also need to assume that the partially colour-blind people are told that there are 15 red and 85 green. Thus the instruction to put all the red balls in bag A means that they have to put 15 balls in bag A. (If this assumption is not made....we'll come back to this later!)
So if they are 80% reliable then they will put 80% of 15 red balls (ie12 red balls) in bag A and therefore 3 green balls to total 15.
So p(red ball from bag A) = 12/15 = 0.8.

Now if we don't make the assumption that they know there should be 15 balls put in bag A, they would "see" 80% of the 15 of the red balls as red (ie 12) and also they would see 20% of the 85 green balls as red (ie 17) so they would put 29 balls in bag A. Since only 12 of them were really red, p(red ball from bag A) = 12/29.

I think I'm liking the second scenario!! Nice question.

3. Thanks alot! I actually came to the same conclusion independently after I posted

I believe the second scenario is what we're after -- they're just given a set of balls (without knowing the ratio of red/green).

4. Originally Posted by scorpion007
Thanks alot! I actually came to the same conclusion independently after I posted

I believe the second scenario is what we're after -- in that they may mistake some green balls for red ones and misplace them. I got the same probability as you.
Cheers.