Balls and Boxes How to address the case where $i\neq j$. Consider the following problem. I cannot seem to prove the case where $i\neq j$ can you please suggest any hints?

Proposition. A ball is in any one of $n$ boxes and is in the $i^{th}$ box with probability $P_i$. If the ball is in box $i$, a search of
  that box will uncover it with probability $\alpha_i$. Show that the
  conditional probability that the ball is in box $j$, given that a
  search of box $i$ did not uncover it, is
  $$\frac{P_j}{(1-\alpha_i)P_i}\text{ }\text{ }i\neq j$$
  $$\frac{(1-\alpha_i)P_i}{1-\alpha_iP_i}\text{ }\text{ }i=j$$

 A: I agree with @TonyHellmuth, the denominator should be $1-\alpha_iP_i$.
Denote the event "a search for box $i$ did not uncover the ball" by $A_i$, then it can be seen that $P(A_i) = 1-\alpha_iP_i$, because $P(A_i^c) = \alpha_iP_i$. In fact, $A_i^c$ is the event that we uncover the ball by a search in box $i$. Denote the event "the ball is in box $j$" by $B_j$. Notice that the statement "If the ball is in box $i$, a search of that box will uncover it with probability $\alpha_i$" means that $P(A_i^c|B_i) = \alpha_i$, thus $$P(A_i^c) = P(B_i\cap A_i^c) +P(B_i^c\cap A_i^c) = P(B_i\cap A_i^c) = P(A_i^c|B_i)P(B_i) =\alpha_iP_i.$$
Here we use the fact that $B_i^c\cap A_i^c =\emptyset$.
Now the conditional probability, by Bayes rule, is given by
$$
\frac{P(A_i\cap B_j)}{P(A_i)} = \frac{P(A_i\cap B_j)}{1-\alpha_iP_i}.
$$
When $i=j$, $P(A_i\cap B_j) = (1-\alpha_i)P_i$. When $i\neq j$, $A_i\cap B_j =B_j$, because $A_i\subset B_j$, thus
$$
P(A_i\cap B_j) = P(B_j) =P_j.
$$ 
In summary we have
$$\frac{P_j}{1-\alpha_iP_i},\ \ j\neq i$$
$$\frac{(1-\alpha_i)P_i}{1-\alpha_iP_i},\ \ j=i$$
It's easy to check that the sum of the conditional probabilities above over $j=1,\cdots,n$ is $1$.
