# 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: $P(A's \text{ flash failing exactly twice}) = \binom{10}{2}(0.9)^8(0.1)^2=.1937$

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?

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

• The question seems to be related to Bayes' rule. For the first part, I would ask two questions: 1) What role would the factor $1/2$ play in terms of Bayes' rule (e.g. prior,posterior,likelihood)? 2) Do the probabilities describe events that are independent or disjoint? – GeoMatt22 Oct 19 '16 at 0:31

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 trials

A: 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