Conditional Probability Exercise - Car Bomb I've been trying to work out this exercise on conditional probability, and it has me completely stumped. The question is as follows:

Consider a garage containing $n$ cars. Exactly one of the cars contains a bomb. The odds of car $i$ containing the bomb equal $p_{i}$. A bomb-sniffing dog goes through the cars. On the condition that car $i$ contains the bomb, then the dog will find it with a probability of $\alpha_{i}$. What are the odds that car $j$ contains the bomb, in the event that the dog did not find a bomb in car $i$?

I've defined my two random variables as


*

*$ B = $ the car bomb, with set $S_{B} = \{1, ..., n\}$

*$ H_{i} = $ the dog's check on car $i$, with set $S_{H_{i}} = \{True, False\}$


So this leaves me with the following information


*

*$\mathbb{P}(B = i) = p_{i}$

*$\mathbb{P}(H_{i} = True|B = i) = \alpha_{i}$


And what I'm left to figure out is $\mathbb{P}(B = j|H_{i} = False).$ Supposedly, the answer should be $\displaystyle\frac{\alpha_{j}p_{j}}{\alpha_{i}}$, but I have no idea how to find it. I've tried applying various probability identities without any success. Perhaps I've misinterpreted the information?
Thanks in advance.
 A: I read your question as saying that the only information we have is that the dog has checked car $i$ and failed to find the bomb. We know nothing about any other checks the dog has done.
Suppose the bomb was in car $i$. The prob the dog failed to find it was $1-\alpha_i$, whereas if the bomb was not in car $i$, the prob the dog failed to find it was 1. So given that the dog failed to find it, the prob that it was not in car $i$ is $$\frac{1-p_i}{1-p_i+(1-\alpha_i)p_i}=\frac{1-p_i}{1-\alpha_i p_i}$$
If the bomb is in car $i$ then the chance it is in car $j$ is 0. If the bomb is not in car $i$ then the chance it is in car $j$ is $\frac{p_j}{1-p_i}$ (because the relative probs for the cars other than $i$ is unchanged). So given that the dog failed to find the bomb in car $i$, the prob that it is in car $j$ is $$\frac{1-p_i}{1-\alpha_ip_i}\frac{p_j}{1-p_i}=\frac{p_j}{1-\alpha_ip_i}$$
This disagrees with your statement that the answer should be $$\frac{\alpha_jp_j}{\alpha_i}$$ But that must be wrong. Because if the dog has not checked car $j$ (or we do not know the results of the check) then how can $\alpha_j$ be relevant?
-------- Added later --------
The OP asked for further explanation of the first formula. This is standard conditional probability. Let $B$ denote the event that the bomb was in car $i$ and $B'$ the event that it was not. Let $D$ denote the event that the dog failed to find it. We use $p(D|B)$ to denote the prob of $D$ given $B$. 
So $p(D|B)=1-\alpha_i$. Similarly $p(D|B')=1$. We also have $p(B)=p_i,p(B')=1-p_i$. Now the standard Bayesian equation is $p(B'|D)p(D)=p(B'\cap D)=p(D|B')p(B')$, so $$p(B'|D)=\frac{p(D|B')p(B')}{p(D)}$$ But $p(D)=p(D|B)p(B)+p(D|B')p(B')$, so as is often the case it is more convenient to write this equation as: $$p(B'|D)=\frac{p(D|B')p(B')}{p(D|B)p(B)+p(D|B')p(B')}$$ which is the first equation above.
--------- Added later still --------
To amplify the second formula. We have $$p(\text{in car }j)=p(\text{in car }j\ |\text{ in car }i)p(\text{in car }i)+p(\text{in car }j\ |\text{ not in car }i)p(\text{not in car }i)$$ But $p(\text{in car }j\ |\text{ in car }i)=0$. So we have $$p(\text{in car }j)=p(\text{in car }j\ |\text{ not in car }i)p(\text{not in car }i)$$ which is the second formula (although the terms are in the other order).
