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A purse contains three fair coins, two double headed and one double tailed. A coin is chosen randomly . What is the probability of picking a double headed coin given that it landed on heads?

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  • $\begingroup$ Unless I am missing something, the double tailed coin cannot land on heads so ... $\endgroup$ – gandalf61 Jul 25 at 14:16
  • $\begingroup$ @gandalf61 The purse contains six coins in total. $\endgroup$ – Magma Jul 25 at 14:22
  • $\begingroup$ @Magma Oh, I see. Three HT coins and two HH coins and one TT coin. That was not how I read the question, but it does make it less trivial ! $\endgroup$ – gandalf61 Jul 25 at 15:11
  • $\begingroup$ @gandalf61 You are right the double tailed coin cannot land on head. So there are two more coins that are double headed. The way you read it was correct indeed. I used the baye's theorem and two tuples of a coin and its outcome. My answer was 1. I felt it was a bit interesting if my calculation is indeed right. Although intuitively anyone will say1 without doing the calculation . $\endgroup$ – Nazifa Taha Jul 25 at 16:26
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Picking a random coin and flipping it will show any of the twelve total faces with equal probability. Since in this case it landed on heads, it's equally likely to show each of the seven heads. How many of those heads belong to double-headed coins?

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Let $A$ denote the event of picking a double-headed coin, and $B$ denote the event of picking a head. We wish to calculate $\mathbb{P}(A|B)$.

By Bayes, $\mathbb{P}(A|B) = \frac{\mathbb{P}(B|A)\mathbb{P}(A)}{\mathbb{P}(B)}$.

Clearly $\mathbb{P}(B|A)=1$, $\mathbb{P}(A) = \frac{2}{6} = \frac{1}{3}$ and $\mathbb{P}(B) = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{6} + \frac{1}{6} + 0 = \frac{7}{12}$.

Therefore your answer is \begin{equation} \boxed{\frac{4}{7}} \end{equation}

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