Exact probability of failure The flash mechanism on camera $A$ fails on $10\%$ of shots, while that of camera $B$ fails on
$5\%$ of shots. The two cameras being identical in appearance, a photographer selects one
at random and takes $10$ indoor shots using the flash.
first we were asked to find the probability of camera $A's$ flash failing exactly twice. Which I found: $P(A's \text{ flash failing exactly twice}) = \binom{10}{2}(0.9)^8(0.1)^2=.1937$  
We are then asked:
Supposing we do not know which camera was selected, find the probability that
the flash mechanism fails exactly twice. What assumptions are you making?   
Do I multiply each cameras chance of failure by $1/2$ and then multiply their probabilities given they are independent?    
We are then asked:
Given that the flash mechanism failed exactly twice, what is the probability that
camera $A$ was selected? I'm not sure where to go with this question.  
Any help would be appreciated, thanks
 A: I will answer the first part:

Supposing we do not know which camera was selected, find the probability that the flash mechanism fails exactly twice

First, lets define two events with the following symbols:
F: The flash fails exactly twice in 10 trialsA: The photographer selects the camera 'A' (so, $\neg$A , or negation of A, means that the photographer selects the camera 'B')
The goal is to find P(F). 
According to the question, we do not know whether A is true or $\neg$A is true, but it is clear that one of them must be true (but not both). Therefore, we can write:
$P(F)=P(F,A)+P(F,\neg A)$
Where the notation P(X,Y) generally means that X and Y occur simultaneously (i.e. X and Y). 
Then, using the rules of conditional probability, we have:
$P(F,A)+P(F,\neg A)=P(F|A)P(A)+P(F|\neg A)P(\neg A)$
Since the photographer takes one camera at random, we may assume:
$P(A)=P(\neg A)=0.5$
And calculating the values of $P(F|A)$ and $P(F|\neg A)$ is easy. Recall that $P(F|A)$ is the probability of failing twice given the camera 'A' was selected. For $P(F|A)$, calculate the value of failing camera 'A' exactly twice, and the for $P(F|\neg A)$, calculate the value of failing camera 'B' exactly twice. Now, you have $P(F)$.
Note that you have already calculated P(F|A):

first we were asked to find the probability of camera A′sA′s flash failing exactly twice. Which I found: P(A′s flash failing exactly twice)=(102)(0.9)8(0.1)2=.1937

