Bayes' Theorem: Detection of bomb in a box 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. 
 A: It's no bad thing you didn't manage to calculate $P(d_1|p_1^c)$ from the information in the question: it can't be done.  
An example probably makes it easier to explain.  Let's say the sensor measures a continuous variable which goes up when a bomb is in the box, and goes down otherwise, but it's a noisy variable.  You can set a threshold and if the signal goes above the threshold, a bomb has been "detected".  With no noise the threshold could be set to be a perfect detector, never being in error.  So the errors are due to noise.
With noise the signal might be pushed to the wrong side of the threshold.  If a high signal is pushed below the threshold that's a missed detection and the probability of that is $1-\alpha_1$.  It depends on the high signal level, the shape of the distribution when a bomb is present and the threshold.
If a low signal (bomb not present) is pushed above the threshold, that's called a "false alarm" for obvious reasons.  The probability of that depends on the low signal level, the shape of the distribution when a bomb is not present and on the threshold.  Note, apart from the threshold, how little that has in common with the other error rate.  It's impossible1 to derive things which depend on the bomb-not-present distribution from $\alpha_1$ which, remember, was calculated based on the bomb being present, but that's what you're trying to do to get $P(d_1|p_1^c)$.
The thing the detection probability $\alpha_1$ and the false alarm probability have in common is a threshold, so you can adjust that to change the tradeoff between the two.  The tradeoff can be visualized with things like the Receiver Operating Characteristic (ROC) curve, but there's no general formula for that curve: it varies from problem to problem and from detector to detector.  
I've used a threshold detector as an example, but it holds for detectors in general, no matter how simple or complex, and all kinds of statistical hypothesis testing.
(1) In general: if you know a lot about a particular sensor and how it's being operated, you may be able to do it.
A: 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.
