# Probability Question

September 3, 2011 5:56 PM Subscribe

Probability Question

A quick question of probability that I can't figure out. For traits A,B,X,Y. If the probability of A and X is 0.9, the probability of A and Y is 0.1, the probability of B and X is 0.75, and the probability of B and Y is 0.25.

If Y, what is the probability of B as opposed to A?

A quick question of probability that I can't figure out. For traits A,B,X,Y. If the probability of A and X is 0.9, the probability of A and Y is 0.1, the probability of B and X is 0.75, and the probability of B and Y is 0.25.

If Y, what is the probability of B as opposed to A?

Yes, A and B are mutually exclusive, X and Y are mutually exclusive

posted by nickhb at 6:07 PM on September 3, 2011

posted by nickhb at 6:07 PM on September 3, 2011

This doesn't seem to make sense to me.

If the probability of A and X is 0.9 and the probability of A and Y is 0.1 and X and Y are mutually exclusive, then the probability of A is 1.

If the probability of B and X is 0.75 and the probability of B and Y is 0.25 and X and Y are mutually exclusive, then the probability of B is 1.

Another way to look at this is that if A and B are mutually exclusive, and X and Y are mutually exclusive, then the combinations A & X + A & Y + B & X + B & Y should cover all the possibilities; that is, they should sum to 1. But according to your statement they sum to 2.

posted by dfan at 6:12 PM on September 3, 2011

If the probability of A and X is 0.9 and the probability of A and Y is 0.1 and X and Y are mutually exclusive, then the probability of A is 1.

If the probability of B and X is 0.75 and the probability of B and Y is 0.25 and X and Y are mutually exclusive, then the probability of B is 1.

Another way to look at this is that if A and B are mutually exclusive, and X and Y are mutually exclusive, then the combinations A & X + A & Y + B & X + B & Y should cover all the possibilities; that is, they should sum to 1. But according to your statement they sum to 2.

posted by dfan at 6:12 PM on September 3, 2011

I hope I'm reading the question right: you've told us X|A, Y|A, X|B, and Y|B. Without knowing the proportion of A to B, this question can't be answered.

posted by Nomyte at 6:12 PM on September 3, 2011

posted by Nomyte at 6:12 PM on September 3, 2011

Once you have a value for the proportion of A to B (i.e. the probability of A, NOT assuming X) then what you are looking for is Bayes' Theorem.

posted by zeptoweasel at 6:16 PM on September 3, 2011

posted by zeptoweasel at 6:16 PM on September 3, 2011

Ok, I'll try that. For those of you wondering, the probability of A is roughly 0.965

posted by nickhb at 6:30 PM on September 3, 2011

posted by nickhb at 6:30 PM on September 3, 2011

Maybe when you said "the probability of A and X" you meant "the probability, given A, of X"? I'm trying to make sense of the question as given.

posted by dfan at 6:36 PM on September 3, 2011

posted by dfan at 6:36 PM on September 3, 2011

Yes, that's correct. Sorry, this is me trying to make sense of some news statistics, and I don't really know how to phrase the question right

posted by nickhb at 6:37 PM on September 3, 2011

posted by nickhb at 6:37 PM on September 3, 2011

You'll understand this more quickly if you pay attention to being a little less sloppy with terminology. You've given probabilities for "A and X" etc but I suspect that what you actually have is probabilities for A given X, or possibly X given A.

The probability of A

The probability of A

Bayes's theorem applies to givens.

posted by flabdablet at 6:37 PM on September 3, 2011

The probability of A

**and**X is the probability that both A and X are true simultaneously.The probability of A

**given**X is the probability that**if**X is true, A is also true.Bayes's theorem applies to givens.

posted by flabdablet at 6:37 PM on September 3, 2011

*Sorry, this is me trying to make sense of some news statistics, and I don't really know how to phrase the question right*

How does the source you're getting this from phrase it?

posted by madcaptenor at 6:39 PM on September 3, 2011

Knowing that P(A) = 0.965, P(B|Y) should be about 0.0836, unless I messed up somewhere.

posted by Nomyte at 6:40 PM on September 3, 2011

posted by Nomyte at 6:40 PM on September 3, 2011

I just went back and checked. A population of animals has two traits with two possible alleles each. Animals are all either A or B, and either X or Y. 3.5% of the Animals are B. If an animal is A, P(X)=0.9 and P(Y)=0.1. If an Animal is B, P(X)=0.75 and P(Y)=0.25 . If you observe an animal to be Y, what are the odds that it is also B?

From what I'm understanding of Baye's Theorem, it would be P = [(0.75)(0.035)]/(0.965)

posted by nickhb at 6:46 PM on September 3, 2011

From what I'm understanding of Baye's Theorem, it would be P = [(0.75)(0.035)]/(0.965)

posted by nickhb at 6:46 PM on September 3, 2011

I don't think you can tell anything without knowing the proportion of A to B (or possibly X and Y) can you?

If there are 1,000 of A and 100 of B then Y will be expected to have 100 A and 25 B, giving a 20% chance of B given Y. If there are 100 of each you'd expect 10 A and 25 B, so over a 70% chance of B, given Y.

posted by Zalzidrax at 6:57 PM on September 3, 2011

If there are 1,000 of A and 100 of B then Y will be expected to have 100 A and 25 B, giving a 20% chance of B given Y. If there are 100 of each you'd expect 10 A and 25 B, so over a 70% chance of B, given Y.

posted by Zalzidrax at 6:57 PM on September 3, 2011

So you're trying to find P(B|Y).

This is P(YB)/P(Y).

P(YB) = P(B) P(Y|B) = (0.035)(0.25) = 0.00875 by assumption.

P(Y) = P(AY) + P(BY) = P(A) P(Y|A) + P(B) P(Y|B) = (0.965)(0.1) + (0.035)(0.25) = 0.0965 + 0.00875 = 0.10525.

The final probability is P(YB)/P(Y) = .00875/.10525 = .0831. (This agrees with Nomyte up to rounding error.)

Incidentally, I usually advise students not to memorize Bayes' theorem but rather to remember how it's derived fom the definition of conditional probability.

posted by madcaptenor at 6:57 PM on September 3, 2011

This is P(YB)/P(Y).

P(YB) = P(B) P(Y|B) = (0.035)(0.25) = 0.00875 by assumption.

P(Y) = P(AY) + P(BY) = P(A) P(Y|A) + P(B) P(Y|B) = (0.965)(0.1) + (0.035)(0.25) = 0.0965 + 0.00875 = 0.10525.

The final probability is P(YB)/P(Y) = .00875/.10525 = .0831. (This agrees with Nomyte up to rounding error.)

Incidentally, I usually advise students not to memorize Bayes' theorem but rather to remember how it's derived fom the definition of conditional probability.

posted by madcaptenor at 6:57 PM on September 3, 2011

*Bayes's theorem applies to givens.*

Yes, but conditionals can be expressed in terms of joint and marginal probabilities, so the distinction is irrelevant:

Pr(X|A) = Pr(XA)/Pr(A) = Pr(A|X)Pr(X)/P(A)

As for the main question, Bayes' rule gives us:

Pr(B|Y) = Pr(Y|B)Pr(B)/Pr(Y)

= Pr(Y|B)Pr(B)/[Pr(Y|A)Pr(A) + Pr(Y|B)Pr(B)]

= ( .25 * .035 ) / ( .1*.965 + .25*.035 )

= .083

Now, the

**odds**of B|Y are the ratio of B|Y to A|Y:

Pr(B|Y)/P(A|Y) = .083/.917 = .091

All of which, I now see, I was too slow to type to be the first to say. Oh, well.

posted by noahpoah at 7:03 PM on September 3, 2011

This thread is closed to new comments.

posted by dfan at 6:06 PM on September 3, 2011