A key is equally likely to be in one of four bins. αi is the probability that you find the key in bin i after looking for it in bin i. (0<= αi <= 1)
You look in bins 2 and 4 and do not find the key. What is the probability that the key is either in bin 2 or 4?
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I'm thinking of Baye's rule
Bi - event that the key is in bin i (i=1,2,3,4)
E - event that you do not find the key in bins 2 and 4 after examining these two bins
then the desired probability is $$ \begin{align*} P(B2\cup B4\mid E)&=P(B2\mid E)+P(B4\mid E)\\&=\frac{P(E\mid B2)P(B2)}{\sum_i P(E\mid Bi)P(Bi)}\\+ \frac{P(E\mid B4)P(B4)}{\sum_i P(E\mid Bi)P(Bi)}\\&=\frac{(1-α2)(1-α4)1/4}{(1-α2)1/4 + 1/4+1/4+1/4}\\+ \frac{(1-α2)(1-α4)1/4}{(1-α4)1/4 + 1/4+1/4+1/4}. \end{align*} $$
Is it close to the correct answer? I'm a little bit confused..
$B_i$ and $E$ make sense as you have defined them (with the exception of the "and" vs "or" case I mentioned in comments).
You are correct in setting up: $P(B_2 \cup B_4 |E)=\frac{P(E|B_2)P(B_2)+P(E|B_4)P(B_4)}{\sum_{i}{}P(E|B_i)P(B_i)}$
$P(E|B_i) = 1$ if $i \ne 2, 4$ since if the key is not in $2$ or $4$ you can't find it there.
$P(E|B_i) = (1-\alpha_i)$ for $i = 2,4$ since if the key is in bin $i$ the probability the $E$ happens only amounts to you not finding it in bin $i$.
$P(B_i) = \frac{1}{4} \ \forall i$
Combining this we get:
$$P(B_2 \cup B_4|E) = \frac{\frac{1}{4}(1-\alpha_2 + 1-\alpha_4)}{\frac{1}{4}(1+(1-\alpha_2)+1+(1-\alpha_4)}=\frac{2-\alpha_2 -\alpha_4}{4-\alpha_2-\alpha_4}$$