Conditional probability exercise - am I thinking right? There is an object, that could be in $3$ places with the probabilities: $$P(A)=0.25, P(B)=0.5, P(C)=0.25 $$
It's surely found in place "A". $P(found|A)=1$, but in the other places it differs: $$P(found|B)=0.9$$ $$P(found|C)=0.5$$ 
What's the probability of it being in "A", if not found in "B" and "C"?
Well, since it is not found, the possibilities are: $$P(not found|C)\cdot P(C)+P(not found|B)\cdot P(B)+P(A)=0.5\cdot 0.25+0.1\cdot 0.5+0.25=0.425$$ Since $P(A)=0.25$, the answer is $0.25/0.425=100/17 \approx 0.588$
The definition goes like this: $$P(X|Y)=\frac{P(X \cap Y)}{P(Y)}$$ 
How could I use this for solving the exercise? Is my solution above any good?
Thanks for your answers!
 A: The correct calculation to make is
$${0.25\over0.25+(0.1\cdot0.5)+(0.5\cdot0.25)}={250\over250+50+125}={10\over17}$$
That is, the denominator takes into account the possibilities that the object is either in $A$ (and will be found there, once you look, since $P(found|A)=1$) or in $B$ but wasn't found there or in $C$ but wasn't found.
The derivation of the denominator in the OP's attempt, $0.825$, can be seen to be wrong by doing a little thought experiment:  assume that $P(found|B)$ and $P(found|C)$ are, like $P(found|A)$, also equal to $1$.  Then the OP's denominator would be
$$P(found|C)\cdot P(C)+P(found|B)\cdot P(B)+P(A)=1\cdot 0.25+1\cdot 0.5+0.25=1$$
which would lead to an answer of $0.25$.  But if the object is certain to be found when looked for, and it's not found in either $B$ or $C$, then it must be in $A$.
A: 
What's the probability of it being in "A", if not found in "B" and "C"? 

Translation: Imagine you have three cupboards, and are looking for the salt.  You have looked in cupboards B and C and not found the salt.   Given that condition, what is the probability of it being in A?
Let $F_A, F_B, F_C$ represent the events: "found in the indicated cupboard", and $A,B,C$ the events "is actually in that cupboard".
The condition is thus: The salt is in A, or the salt was in B but not found there, or the salt was in C but not found there. $\;F_B^\complement\cap F_C^\complement = A\cup (B\cap F_B^\complement)\cup (C\cap F_C^\complement)$

$$\begin{align}
\mathsf P(A\mid F_B^\complement\cap F_C^\complement) 
& = \dfrac{\mathsf P(A)}{\mathsf P(A)+\mathsf P(B)(1-\mathsf P(F_B\mid B))+\mathsf P(C)(1-\mathsf P( F_C\mid C))}
\\[1ex]
& = \dfrac{0.25}{0.25+0.5(1-0.9)+0.25(1-0.5)}
\\[1ex]
& = \dfrac{10}{17}
\end{align}$$
