Bayes Probability problem on double headed coins The question says that you have five coins, two of which are double-headed, one is double-tailed, and the
other two are normal. You shut your eyes, pick a coin at random, and toss it. The first toss is showing a head. 


*

*You shut your eyes again and toss the coin again. What is the probability that the
lower face is a head?

*You open your eyes and see that the coin is showing heads. What is the probability
that the lower face is a head?
Appreciate if anyone can walk me thru the steps on how to solve these two problems. Thanks!
By drawing a tree diagram, it is clear that the answer for the first problem is 5/6 and 4/5 for the second problem. But I am not too sure how to solve the problem using bayes theorem... 
 A: Hint:
There would be $20=5\times 2 \times 2$ equally likely cases of coins and orientations for the two tosses
For question 1, you can eliminate those showing tails in the first toss, and count the lower faces for the second toss
For question 2, you can eliminate those showing tails in the first toss or in the second toss, and count the lower faces for the second toss
Added
As you have found, that approach will get you to $\frac{10}{12}=\frac{5}{6}$ for question 1 and $\frac{8}{10}=\frac{4}{5}$ for question 2.  If you want an more explicitly Bayes theorem approach, define the events  


*

*$A$ is the event that the second toss has a lower face of heads

*$B$ is the event that the first toss has a higher face of heads

*$C$ is the event that the second toss has a higher face of heads

*$D$ is the event that the coin is double headed

*$E$ is the event that the coin is double tailed

*$F$ is the event that the coin is normal


For question 1 you want $$\Pr(A \mid B) = \dfrac{\Pr(A \cap B)}{\Pr(B)} = \dfrac{\Pr(A \cap B \cap D)+\Pr(A \cap B \cap E)+\Pr(A \cap B \cap F)}{\Pr(B \cap D) + \Pr(B \cap E)+ \Pr(B \cap F)} $$  
$$= \dfrac{\Pr(A \cap B \mid D)\Pr(D) +\Pr(A \cap B  \mid E)\Pr(E)+\Pr(A \cap B  \mid F)\Pr(F)}{\Pr(B  \mid D)\Pr(D) + \Pr(B  \mid E)\Pr(E)+ \Pr(B  \mid F)\Pr(F)} $$
$$= \dfrac{1 \times \frac25 +0 \times \frac15 +\frac14\times \frac25}{1 \times \frac25 +0 \times \frac15 +\frac12\times \frac25} = \dfrac{\frac{1}{2}}{\frac{3}{5}} = \dfrac56$$
For question 2 you want something similar which ends up with $$\Pr(A \mid B, C) = \dfrac{\Pr(A \cap B\cap C)}{\Pr(B\cap C)}= \cdots $$ $$= \dfrac{1 \times \frac25 +0 \times \frac15 +0\times \frac25}{1 \times \frac25 +0 \times \frac15 +\frac14\times \frac25}= \dfrac{\frac{2}{5}}{\frac{1}{2}} = \dfrac45$$
