There is a bomb that is equally likely to be in any one of three different boxes. Let $α_i$ be the probability that that the bomb will be found upon making a quick examination ( detection ) of box i if the bomb is, in fact, in box i, i =1,2,3. (We may have $α_i$ <1.) Suppose you examine box 1 and do not detect the bomb. What is the probability that the bomb is in box 1?
My approach is as follows:- Let $d_1$ be the event that bomb is detected in box 1 and $p_1$ be the event that it is, indeed, present in the box 1. Hence, I think, $P(d_1) = \alpha_1$ and $P({p_1}) = \frac{1}{3}$
Then, by conditional probability:-
P(presence of the bomb when it wasn't detected ) i.e P($p_1 | {d_1}^c$) = $\frac{P(p_1{d_1}^c)}{P({d_1}^c)}$
where :-
${d_1}^c$ is the complement event of the event ${d_1}$ and $P(p_1{d_1}^c)$ is probability that ${d_1}^c$ and $p_1$ occur together.
Now, $\frac{P(p_1{d_1}^c)}{P({d_1}^c)}$ = $\frac{P(p_1{d_1}^c)}{1-P({d_1})}$ = $\frac{P(p_1)*P({d_1}^c|{p_1})}{1-P({d_1})}$ = $\frac{P(p_1)*(1-P({d_1}|{p_1}))}{1-P({d_1})}$ = $\frac{\frac{1}{3}*({1-}\alpha_1)}{1-P({d_1})}$ = $\frac{1}{3}*\frac{({1-}\alpha_1)}{1-P({d_1})}$-----------------(A)
$P(d_1) = P(d_1p_1) + P(d_1{p_1}^c)$ = $ P(p_1)*P(d_1|p_1) + P({p_1}^c)*P(d_1|{p_1}^c) $ = $\frac{1}{3}*\alpha_1 + \frac{2}{3}*P(d_1|{p_1}^c)$
I am not able to calculate $P(d_1|{p_1}^c)$. Can you please some help ( but not the complete solution ). If you have a better idea, then please share it with me but please do not give the complete solution.
Here goes:
$\alpha_i$, where $i = 1, 2, 3$ is the probability of the bomb present given that it was detected.
$P(p_i)$ is the probability of the bomb actually present.
You defined that $P(d_1)$ is the probability of the bomb being detected.
The key problem in your answer is that $P(d_1)$, defined as the probability of the bomb being detected, is not $\alpha_1$. Also, you cannot take for granted that $P(p_1) = \frac 1 3$.
Hence, in your case
$$\alpha_1 = P(p_1 | d_1)$$ To find $P(d_1^c) $ (I assume in terms of $\alpha_1$ and $p_1$) Use the variables provided above. Try again.
Good luck.