Conditional probabilities of spinning spinners I want to make sure I have the correct understanding of conditional probability.
Consider the following scenario:
You have three spinners, each with 6 equally sized regions, that are marked with a letter:
Spinner 1: A, B, C, D, E, F
Spinner 2: C, D, D, E, E, F
Spinner 3: A, A, A, C, C, C
Lets say you pick a spinner at random and spin it without looking to see which spinner it is and you're told it lands on C. If you spin the same spinner again, what is the probability you get A?
Here's how I'm thinking about it: first, use conditional probability formula to figure out which spinner you spun:
Probability of spinner 1 = P(C|spinner 1) = 1/6
Probability of spinner 2 = P(C|spinner 2) = 1/6
Probability of spinner 3 = P(C|spinner 3) = 2/3
Then, to get the probability of getting A on the second spin, take (P spinner 1)(P spinner 1 yielding A) + (P spinner 2)(P spinner 2 yielding A) + (P spinner 3)(P spinner 3 yielding A)... (1/6)(1/6) + (1/6)(0) + (2/3)(1/2) = 1/36 + 1/3... 13/36
Does this look correct? 
Thanks!
 A: I will assume that you choose which spinner to spin uniformly at random.  I will abbreviate notation as $S_1, S_2, S_3$ representing the events "picked first spinner", "picked second spinner", and "picked third spinner" respectively.
We are given then the following information:  $P(S_1)=P(S_2)=P(S_3)=\frac{1}{3}, P(C|S_1)=P(C|S_2)=\frac{1}{6}, P(C|S_3)=\frac{1}{2}$
We can now calculate $P(C)$ as $P(C)=P(C\cap(S_1\cup S_2\cup S_3))=P(C\cap S_1)+P(C\cap S_2)+P(C\cap S_3)$ since $S_1,S_2,S_3$ forms a partition of our sample space.
Continuing by applying the multiplication principle:
$=P(S_1)P(C|S_1)+P(S_2)P(C|S_2)+P(S_3)P(C|S_3) = \frac{1}{3}\frac{1}{6}+\frac{1}{3}\frac{1}{6}+\frac{1}{3}\frac{1}{2}=\frac{5}{18}$
(This calculation could have been figured out immediately by noting that each of the 18 regions are equally likely)
Applying Bayes' Theorem, you get $P(S_3|C) = \frac{P(C|S_3)P(S_3)}{P(C)} = \frac{\frac{1}{2}\frac{1}{3}}{\frac{5}{18}}=\frac{3}{5}$
(again, this calculation could have been figured out immediately by noting that of the five $C$'s across all of the boards, since each are in fact equally likely, three of the five are on spinner 3)

Similarly, we calculate $P(S_1|C)=P(S_2|C)=\frac{1}{5}$
Now, let us let $C_1$ represent the event that the first spin was a $C$, and let $A_2$ represent the event that the second spin was an $A$.  We ask the question, what is $P(A_2|C_1)$.
We can break this apart via multiplication principle as well as $Pr(A_2|S_1\cap C_1)Pr(S_1|C_1)+Pr(A_2|S_2\cap C_1)Pr(S_2|C_1)+Pr(A_2|S_3\cap C_1)Pr(S_3|C_1)$
$ = \frac{1}{6}\frac{1}{5}+\frac{0}{6}\frac{1}{5}+\frac{3}{6}\frac{3}{5}=\frac{1}{3}$
