Conditional probability, Baye's rule, prisoner Ural / Siberia + coat exercise I am currently statistics and probability course. One of the questions in the textbook is following:

A prisoner will be sent to either Urals or Siberia, but he does not
  know where. He knows, that the probability is 0.8 that he will be sent
  to Siberia. He also knows the probabilities of prisoners wearing a
  coat - 0.5 in Siberia and 0.7 in Urals.
When he arrives to the exile, the first person he sees is not wearing
  a coat.
  
  
*
  
*What is the probability he is in Siberia? 
  
*Given previous information, he sees another one, not wearing a coat. What is probability now, that he is in Siberia? 
  
*Would it change the probability if he saw both previous people at once?
  

I believe the answer for the 1. is following:
$$\mathcal P(S|\bar C)={\mathcal P(S)\times\mathcal P(\bar C|S) \over \mathcal P(\bar C)}={0.8\times0.5\over 1-(0.8\times0.5+0.2\times0.7)}=\frac{20}{23}\dot=\,0.8696$$
However, I have no idea how to move further on 2. or 3. Could you please help me? Thank You!
EDIT
The professor quickly responded with the solution for 2. :
$$\frac{0.8696\times0.5}{0.8596\times0.5+0.1304\times0.3}=0.9174$$
From this, I've found out that the probabilities can be "chained", and decoded the underlining formula as:
$$\mathcal P(S|\bar{C_2})=\frac{\mathcal P(S|\bar C)\times\mathcal P(\bar C|S)}{\mathcal P(S|\bar C)\times\mathcal P(\bar C|S)+\mathcal P(\bar S|\bar C)\times\mathcal P(\bar C|\bar S)}$$
However, since 0.5 is the probability for $\mathcal P(\bar C|S)$ and also for $\mathcal P(C|S)$, I need a clarification, if my decoded formula is right. Thank You.
 A: 
A prisoner will be sent to either Urals or Siberia, but he does not know where. He knows, that the probability is 0.8 that he will be sent to Siberia. He also knows the probabilities of prisoners wearing a coat - 0.5 in Siberia and 0.7 in Urals.
When he arrives to the exile, the first person he sees is not wearing a coat.

*

*What is the probability he is in Siberia?

*Given previous information, he sees another one, not wearing a coat. What is probability now, that he is in Siberia?

*Would it change the probability if he saw both previous people at once?


$$\rm P(S)=\frac{0.8*0.5}{0.8*0.5+0.2*0.3}=\frac{20}{23}=0.8695\cdots$$
$$\rm P(S_2)=\frac{0.8*0.5*0.5}{0.8*0.5*0.5+0.2*0.3*0.3}=\frac{100}{109}=0.9174\cdots$$
No It won't.

Edit:
$$\rm P(S|(\bar C_1\cap\bar C_2))=\frac{P(S\cap\bar C_1\cap\bar C_2)}{P(\bar C_1\cap\bar C_2)}$$
Now since $\rm P(S\cup U)=1$
$$\rm P(S|(\bar C_1\cap\bar C_2))=\frac{P(S\cap\bar C_1\cap\bar C_2)}{P((S\cup U)\cap\bar C_1\cap\bar C_2)}=\frac{P(S\cap\bar C_1\cap\bar C_2)}{P((S\cap\bar C_1\cap\bar C_2)\cup(U\cap\bar C_1\cap\bar C_2))}$$
And Now, $\rm P(S\cup U)=P(S)+P(U)-P(S\cap U)$, but $\rm P(S\cap U)=0$, So:
$$\rm P(S|(\bar C_1\cap\bar C_2))=\frac{P(S\cap\bar C_1\cap\bar C_2)}{P(S\cap \bar C_1\cap\bar C_2)+P(U\cap\bar C_1\cap\bar C_2)}$$
