0
$\begingroup$

Assume you are searching an area for a missing item. The item is static and does not move.

The probability of the item being in the area is 0.6; the expected search effectiveness is 0.4. Thus, the probability of detecting the item in a given search is 0.6 * 0.4 = 0.24.

Assuming each search is independent of the one before, how do you calculate the combined probability of detection for two searches? Intuitively, the more times you search, the better your chance of finding the item, but in some cases the searches might closely overlap, and in others there may be little overlap.

$\endgroup$
0
$\begingroup$

Maybe a simpler way to think about this is to think about the chance that you search twice and don't find the item. We know that the probability of finding the item in a single search is 0.24 thus the probability we don't find it is (1 - 0.24). So the probability of detection for two searches (let's call it event $A$) would be:

$P(A) = 1.0 - {(1.0 - 0.24)}^2 = 0.4224$

$\endgroup$
0
$\begingroup$

I interpret the problem statement in a way that "independent" means: If the item is surely in the area, then the probability of both searches finding it is the product of the individual probabilities. These probabilities are what you called search effectiveness. It follows that two independent searches with effectiveness $0.4$ have a combined effectiveness of $1-(1-0.4)^2=0.64$. Thus the probability of the item being found by at least one of the searches is $0.6\cdot 0.64 = 0.384$.

It may be worth noting, however, that the events "item is found by first search" and "item is found by second search" are not independent (namely, $1-(1-0.24)^2\ne 0.384$). Indeed, if this were the case, we could just perform enough independent searches and retrieve the item with a probability much higher than the probability of its actual presence - which is absurd.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.