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BucketOfFish
05-03-2010, 01:32 PM
I posted this question on Y! Answers a while ago and no one answered it :) I thought I'd try my luck here instead.

A scientist performs an experiment to test two hypothesis. If hypothesis A is correct, there is a 20% chance that the experiment succeeds. If hypothesis B is correct, there is a 100% chance that the experiment succeeds. The two hypothesis are mutually exclusive, and one of them is definitely correct. If the experiment does in fact succeed, what are the chances that hypothesis A is the correct one?

I don't know if I did this correctly, but I did

P(B)=P(A')
P(C|A)=.2
P(C|B)=1

P(A|C)=P(AnC)/P(C)
=P(C|A)*P(A)/P(C)

P(C)=P(C|A)*P(A)+P(C|B)*P(B)

P(A|C)=(P(C|A)*P(A))/(P(C|A)*P(A)+P(C|B)*P(B))
=.2P(A)/(1-.8P(A))

Here's where I get mixed up. I don't know whether or not to say that hypothesis A and B are initially equally likely, and set P(A)=P(B)=.5. If I do, then

P(A|C)=.2*.5/(1-.8*.5)
=1/6

But I don't know whether or not I can do that. It seems like a really arbitrary thing to do. Could somebody out there help me please? This is really bothering me. Thanks!

fed1
05-03-2010, 04:15 PM
The question as it is phrased is non sensical

Pr( hypothesis A is correct given any condition ) =
1 if it is correct, 0 if it is not.

The true state of nature is not the outcome of an experiment. (frequentist opinion)

Statisticians usually talk about liklihood in this situationn.

one could take the view that hypothesis a is a parameter in a family and find the liklihood associated with it.

Let f1 = bernoulli(20%), f2 = bernoulli(1).

X is the outcome of your trial

X ~ bernoulli( F1(theta) + f2(1 - theta) ).

theta = 1 if hyp A correct.

and so forth.