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Question : I have a fair coin and a two-headed coin. I choose one of the two coins randomly with equal probability and flip it. Given that the flip was heads, what is the probability that I flipped the two-headed coin

Probability of picking a coin is 1/2. probability of head in two headed coin is 1. enter image description here

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  • $\begingroup$ hint...you are looking for a conditional probability $\endgroup$ – David Quinn Apr 8 '18 at 16:04
  • $\begingroup$ Not quite right yet. Finish the tree showing all four leaves with probability $1/4$ for each. How many heads? Of those, how many with the unfair coin? $\endgroup$ – Ethan Bolker Apr 8 '18 at 17:15
  • $\begingroup$ @Ethan Bolker There will be three because one coin is double head $\endgroup$ – user547750 Apr 8 '18 at 17:17
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If $DH$ stands for double-headed coin, and $H$ for Heads, you need $$p(DH|H)=\frac{p(DH\cap H)}{p(H)}$$ $$=\frac{\frac 12\times 1}{\frac 12\times 1+\frac 12\times\frac 12}$$ $$=\frac 23$$

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Define the following events:

  • $E = \{\text{flip head}\}$
  • $E_1 = \{\text{pick the two headed coin}\}$
  • $E_2 = \{\text{pick the fair coin}\}$

We want to compute the probability of $E_1$, given $E$.

Note that $E_1$ and $E_2$ is a partition of the sample space. By the Baye's rule $$ \mathbb{P}(E_1 | E) = \frac{\mathbb{P}(E | E_1) \mathbb{P}(E_1)}{\mathbb{P}(E | E_1) \mathbb{P}(E_1) + \mathbb{P}(E | E_2) \mathbb{P}(E_2)} $$

It is easy to see that

  • $\mathbb{P}(E | E_1) \mathbb{P}(E_1) = \frac{1}{2}$
  • $\mathbb{P}(E | E_2) \mathbb{P}(E_2) = \frac{1}{4}$

Therefore, $$ \mathbb{P}(E_1 | E) = \frac{2}{3} $$

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