A box has three coins. One has two heads, another two tails and the last is a fair coin. I am stuck on this question:
A box has three coins. One has two heads, one has two tails, and the other is a fair coin with one head and one tail.  A coin is chosen at random, and comes up head.  
a) What is the probability that the coin chosen is the two headed coin
b) What is the probability that if it is thrown another time it will come up heads
c) What is the probability that the coin chosen is the two headed coin, supposing that the coin is thrown a second time and comes up heads again
I have solved part a, and the answer is $\frac 23$.  
However I am stuck on parts b and c.  This is my thought process:
part b: P(Heads on the second throw) = P($H_2$|$H_1$) $\cdot$ P(fair coin) + P($H_2$|$H_1$) $\cdot$ P(coin with two heads) = $\frac 12$ $\cdot$ $\frac 13$ + 1 $\cdot$ $\frac 13$ = $\frac 12$, and I'm not sure how to even start with part c.  
The answer to part b should be $\frac 56$, but can anyone explain why?  Thanks.

Your answer to part (a) is incorrect. If you pull out a random coin and flip, you have six scenarios:


*

*Head 1 of 2-headed coin

*Head 2 of 2-headed coin

*Head of fair coin

*Tail of fair coin

*Tail 1 of 2-tailed coin

*Tail 2 of 2-tailed coin


3 of those are "heads", and 2 of those 3 correspond to the 2 headed coin. Thus, the answer to part (a) is $\frac 23$
Part (b): Probability of getting double headed * getting head from that double headed +  Probability of getting fair * getting head from that fair coin
$$\frac 23 \cdot 1+\frac 13 \cdot \frac 12=\frac 56$$
Part (c): Now you have $3\cdot 2\cdot 2=12$ possiblities; check all of them and see which of the ones involving two heads involve the double headed coin.
You can also look at Bayes's Theorem; it covers problems like this one.
