Conditional Probability given 3 events Say I have a pack of 60 batteries. There is a 95% probability that all of them work, a 3.5% probability that 1 is defective, and 1.5% probability that 2 are defective. If I choose 2 from the pack, and the first I test is defective, what is the probability that the second is also defective? 
I know I am looking for the probability of 2 defective batteries given one defective, so a conditional probability, however I can't figure out how to solve it. All of the formulas I have for solving these involve the probability of an intersection, a reversal of the condition, or compliments. I don't seem to have any. Can anyone help me start this problem? Am I looking too far into an easy problem?
 A: Let $A$ be the event that the first battery you pull is defective.  Let $B$ be the event that the second battery you pull is defective.
Let $X$ be the event that the pack contains one defective battery total and $Y$ the event that it has two defective batteries total.
We are asked to find $Pr(B\mid A)$
We have:
$$Pr(B\mid A) = \dfrac{Pr(B\cap A)}{Pr(A)} = \dfrac{Pr(B\cap A\cap X)+Pr(B\cap A\cap Y)}{Pr(A\cap X)+Pr(A\cap Y)} = \dfrac{Pr(B\mid A\cap X)Pr(A\mid X)Pr(X)+Pr(B\mid A\cap Y)Pr(A\mid Y)Pr(Y)}{Pr(A\mid X)Pr(X)+Pr(A\mid Y)Pr(Y)}$$
Can you continue from here?

 $\dfrac{0+\frac{1}{59}\cdot \frac{2}{60}\cdot 0.015}{\frac{1}{60}\cdot 0.035+\frac{2}{60}\cdot 0.015}\approx 0.0078227$

A: JMoravitz's answer is perfect, and this is not a criticism of it in any way.  It's just a slightly different (though entirely equivalent) way of thinking about it that I find easier.
Using the same notation as in that answer, let us compute the a posteriori probabilities of $X$ and $Y$, after the first battery turns out to be defective.  We have $$\Pr(A)=\Pr(A|X)\Pr(X)+\Pr(A|Y)\Pr(Y)={.035\over60}+{2\cdot.0105\over60}={.065\over60}$$ Then $$\Pr(X|A)={\Pr(X\cap A)\over\Pr(A)}={\Pr(A|X)\Pr(X)\over\Pr(A)}={.035/60\over.065/60}={7\over13}$$ and similarly (or because it it the only possibility left) $$\Pr(Y|A)={6\over13}$$
So now the situation is that there are $59$ batteries in a box with at most one defective battery, and the probability that there is a defective battery is ${6\over13}$.  Clearly, the probability of drawing a defective battery is $${6\over13\cdot59},$$ the same answer JMoravitz got.  
