Question is
There are 8 fair coins (C0)and 3 unfair coins(C1) (80% chance get head) tosses it n times without looking at it , and reports:
n=2, got 1 T and 1 H , don't know the order. what is the probability the coin is fair coin?
My solution:
Have two situations , either to be HT or TH so calculate Pr[ C0 | HT] = (Pr[HT | C0] * Pr[C0])/Pr[HT]=0.807 using bayes theorem.
Similarly , Pr[ C0 | TH] = ( Pr [ TH | C0] * Pr[C0])/Pr[TH]=0.807
So the question is do I add up these situation or get product from them to calculate the probability ? I feel like because probability values were expressed in the range 0 ≤ P≤ 1,
but I am not so sure.
No, you do not add the probabilities together. You should be able to see this as it would lead to a number greater than $1$, which cannot be a probability, as you have said.
Your calculation should be $$\Pr[ C_0 | HT \text{ or } TH] = \dfrac{\Pr[HT \text{ or } TH | C_0] \;\Pr[C_0]}{\Pr[HT \text{ or } TH]}$$
and you will find it gives the same value of about $0.807$, since $\Pr[HT \text{ or } TH | C_0]=2\Pr[HT | C_0]$ and $\Pr[HT \text{ or } TH]=2\Pr[HT ]$.