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Three coins but into hat. Coin 1 shows heads on both sides, coin 2 is a fair coin, and coin 3 has a .75 probability of landing heads...one coin is randomly selected and flipped. The coin lands heads side up. What is the probability that the other side is also heads? ... My friend and I are debating the answer to this question. One of us thinks it is 12/27, and the other 1/3. Who is correct?

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This is my take on it:

Provided that heads show up, what is the probability that the coin it showed up on was coin 1?

So let's say, B=Probability that heads show up, and, A=Probability that you chose coin 1.

So we want to calculate the conditional probability of $P(A|B)$.

According to Baye's theorem we get: $$P(A|B)=\frac{P(A)P(B|A)}{P(B)}$$ $$P(A)=\frac{1}{3}$$ $$P(B)=\frac{1+0.5+0.75}{3}=\frac{3}{4}$$ $$ P(B|A)=1$$ $$\therefore P(A|B)=\frac{\frac{1}{3}}{\frac{3}{4}}=\frac{4}{9}(=\frac{12}{27})$$

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  • $\begingroup$ Yes! I agree. This was actually an exam question, and as I left my friends and I disagreed on the answer. My professor sided with my friend, but I did it out again and still got 12/27. Thanks for the response $\endgroup$ – mm8511 Oct 6 '14 at 23:08

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