# fair and double head coin

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.

• hint...you are looking for a conditional probability – David Quinn Apr 8 '18 at 16:04
• 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? – Ethan Bolker Apr 8 '18 at 17:15
• @Ethan Bolker There will be three because one coin is double head – user547750 Apr 8 '18 at 17:17

If $DH$ stands for double-headed coin, and $H$ for Heads, you need $$p(DH|H)=\frac{p(DH\cap H)}{p(H)}$$ $$=\frac{\frac 12\times 1}{\frac 12\times 1+\frac 12\times\frac 12}$$ $$=\frac 23$$

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