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Can someone help me with a basic Bayesian Search Theory question? I'm not sure how to approach these questions and after reading up about the Bayesian search theory I'm still wondering how I should approach these questions. The question is:

A ball is in one box out of $n$. For $i = 1,..,n$ the ball is in the $i$-th box with probability $p_i$. For each value of $i$, there is a probability $\alpha _i$ that a search of the $i$-th box will reveal the ball if it is actually in that box.

What is the probability that the ball is in box 1 given that a search of box 3 fails to reveal the ball? Also, what is the probability the ball is in box 1 given that a search of box 1 fails to reveal the ball?

Any advice on how to tackle these sort of questions will be appreciated.

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What you want in the first question is $$ Pr(\text{Ball in Box 1}|\text{Ball not found in Box 3})=\frac{Pr(\text{Ball in Box 1}, \text{Ball not found in Box 3})}{Pr(\text{Ball not found in Box 3})} $$

From the phrasing of the question it seems like there is no chance of "false positives". So if the ball is in Box 1, then the ball will never be found in box 3. So the numerator reduces to $$ Pr(\text{Ball in Box 1}, \text{Ball not found in Box 3})=Pr(\text{Ball in Box 1})=p_1 $$ For the denominator, this can happen in two ways. To not find the ball in box 3, it is required that 1) the ball is not in box 3, or 2) the ball is in box 3, but the search fails. So then this is: $$ Pr(\text{Ball not found in Box 3})=(1-p_3)+p_3\alpha_3=1-(1-\alpha_3)p_3. $$

Putting this together, the result is

$$ Pr(\text{Ball in Box 1}|\text{Ball not found in Box 3})=\frac{p_1}{1-(1-\alpha_3)p_3}. $$

For the second question, I think you do it using similar reasoning. Hope this helps.

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  • $\begingroup$ Thanks. When you use the notation Pr(A,B) - are you using this comma like an intersection? $\endgroup$ – Ronald Walker Oct 17 '13 at 10:12
  • $\begingroup$ @RonaldWalker Precisely! $\endgroup$ – hejseb Oct 17 '13 at 10:37

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