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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.

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Did you craft the question yourself? I think it's a poorly phrased question. I can offer ways to improve the question, though. –  bryansis2010 Feb 26 '13 at 13:45
    
I apologize for the inconvenience. Please free to edit the question to make it more understandable. By the way, what was I unable to convey properly? –  kusur Feb 26 '13 at 13:49
    
I have read your formulae. They all look correct. However, I think it was a wrong context. The probabilities are represented wrongly. Furthermore, it is hard to imagine the idea where "probability a bomb exists" given "you actually saw a bomb" in a box...it's almost 1. If you would like to, I will change the question to make it easier to fit a practice question for Bayes Theorem. –  bryansis2010 Feb 26 '13 at 13:54
    
The bomb is supposed to be detected, not seen. There might be issues with the detector that it is doing so. It might happen that the bomb is pressure sensitive and opening the box might trigger. So, the bomb is to be detected and it might happen that the detector was unable to detect it. Think of the medical cases where a patient is said to free of a disease when , in fact, he has that disease. –  kusur Feb 26 '13 at 14:06
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2 Answers

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.

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Thanks for the answer. So, from your answer I take it that it not possible to calculate the probability that is required by me. I can say that it is not possible to calculate $P(d_1|{p_1}^c)$ without knowing the threshold and the distribution. Is that right? Also, does you theory also apply if replace bombs with something more safe( say letters ) and boxes with envelopes? If yes, then what will be threshold ( in that case ) ? –  kusur Mar 2 '13 at 18:41
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It's applicable to all kinds of detect/don't detect scenarios. A bit of googling turned up this for explosives detection: link. It gets to Bayes Theorem on page 23. And there's a very similar set of curves and calculations here for medical statistics: [link] (medcalc.org/manual/roc-curves.php). I'm not an expert in either field and I only skim-read them but the statistics looks sound in both. The diagrams of ROC curves and thresholds are fairly standard. –  Gruntled Mar 3 '13 at 13:09
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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.

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In the problem itself it is mentioned that the bomb is equally likely to be present in any of the three boxes. So,I think that$P(p_i)=\frac{1}{3}$. Also, you are right that $\alpha_i$ is the probability that the bomb is detected, provided it is present in the the box i ( i=1,2,3). $P(d_i)$ is indeed denoting just the detection. So ,when I say $P(d_i|p_i)$, I mean detection when it was present and hence, $P(d_i|p_i)$ = $\alpha_i$. –  kusur Feb 26 '13 at 16:39
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