conditional probability or Bayes' theorem? 


My try :
we can deal three different cases of removing two biased coin and 1 unbiased coin.
Individual probability is known $$\frac9{10} , \frac2{25},\frac1{50}$$
After that total is 49 as one is removed.
now what to do conditional probability or Bayes' theorem ? i dont know how to implement ?
 A: Here's how I think about these problems.  There are $50$ coins with $2$ sides each, or $100$ possible outcomes, each equally likely.  Of those possible outcomes, $53$ result in heads and $47$ result in tails.
Out of the $53$ possibilities that result in heads, $8$ of them come from a two-headed coin and the other $45$ come from fair coins.  So if your result is heads, the probability of a two-headed coin is $\frac{8}{53}$ and the probability of a fair coin is $\frac{45}{53}.$  If the result is tails, the probability of a fair coin is $\frac{45}{47}$.
A: There are four possible combinations:


*

*Double-headed and shows heads with probability $\frac{4}{50}\times 1= 0.08$

*Double-tailed and shows tails with probability $\frac{1}{50}\times 1 =0.02 $

*Fair coin and shows heads with probability $\frac{45}{50}\times\frac12 = 0.45$

*Fair coin and shows tails with probability $\frac{45}{50}\times\frac12 = 0.45$
so using ${P(A,B)}=P(A \mid B) \,P(B)$, i.e. $P(A \mid B) =\dfrac{P(A,B)}{P(B)}$, and $P(B)= P(A,B)+P(A^c,B)$, you get 
(a) Given it shows heads, the conditional probability of being double-headed is $\frac{0.08}{0.08+0.45}$
(b) Given it shows heads, the conditional probability of being fair is $\frac{0.45}{0.08+0.45}$
(c) Given it shows tails, the conditional probability of being fair is $\frac{0.45}{0.02+0.45}$ 
