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September 2, 2014

September 2, 2014

Posted by **akshay** on Friday, August 31, 2012 at 11:25am.

- probability -
**David Q/R**, Friday, August 31, 2012 at 4:13pmOkay - this is the first time I've ever tried to use Bayes Theorem...

P(A|B) = P(B|A).P(A) / Sum(P(B|Aj).P(Aj)

Let A be the event "All five balls are white".

Let Aj be the event "There are exactly j white balls". Let B be the event "I draw two white balls".

Then P(A|B) = P(All five balls are white, given that I draw two white balls).

P(B|A) = P(I draw two white balls, given that all five balls are white), and this obviously equals 1.

P(B|A0) = P(I draw two white balls, given that no balls are white) = 0.

P(B|A1) = P(I draw two white balls, given that exactly one is white) = 0.

P(B|A2) = P(I draw two white balls, given that exactly two are white) = (2/5)x(1/4) = (1/10).

P(B|A3) = P(I draw two white balls, given that exactly three are white) = (3/5)x(2/4) = (3/10).

P(B|A4) = P(I draw two white balls, given that exactly four are white) = (4/5)x(3/4) = (6/10).

P(B|A5) = P(I draw two white balls, given that all five are white) = 1.

Now, here's the subjective bit - because Bayes Theorem is all about subjectivity. I have no idea a priori how many white balls there are in the bag, so all possibilities are equally likely as far as I'm concerned. That means that P(A0) = P(A1) = P(A2) = P(A3) = P(A4) = P(A5) = (1/6). So in the original equation, all of the P(Aj)s in the numerator and the denominator are going to cancel out.

That leaves P(B|A) in the numerator, and (P(B|A0) + P(B|A1) + P(B|A2) + P(B|A3) + P(B|A4) + P(B|A5)) in the denominator. The numerator = P(B|A) = 1 (we worked it out above). The denominator is the sum of all those other fractions, which is 0 + 0 + (1/10) + (3/10) + (6/10) + 1 = (10/10) + 1 = 2. So the answer is (1/2), i.e. exactly one half.

Do you reckon I've got it right?

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