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I have a scenario of 3 disparate software utilities (virus scanners - all latest versions). Their hit ratios are excellent, but occasionally 1 of them misses a virus.

0 = no virus, 1 = virus discovered

At some point one of them will fail to detect, i.e. $$(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0)...(0,0,1)$$

However,in an even more rare scenario 2 of them might fail to detect $$(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0),(0,0,0)...(0,0,1)..., ...(1,0,1)$$

Now say I was to use 5 virus scanners, it would be an even greater rare scenario for 3 to fail detection. The question I have is, how can I express the above scenarios in a single Binomial Coefficient ?

mathreadler
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alortimor
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1 Answers1

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Without further information, you have to assume the probability each scanner misses a virus is the same at $p$ and the chances of each scanner missing a virus are independent. These are terrible assumptions. The chance that two of three will miss a virus is then ${3 \choose 2}p^2(1-p)$ and the chance that all three will miss it is ${3 \choose 3}p^3$. You can do the same for fives scanners and $3,4,5$ missing it. If $p$ is rather small, the fact that three of five has a leading term of $p^3$ while the two of three case is proportional to $p^2$ guarantees that your are better off with five. Of course, if one scanner misses a virus it could be the virus is too new for any to catch and the independence assumption fails. Then it doesn't matter how many you have.

Ross Millikan
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  • Thanks for your reply, does that mean that the chances are 1/(2^5) , i.e. given 5 virus scanners the chance of all 5 failing is "1 in 2^5" ? But in the real world, a new virus will fail all 5, if the scanners aren't updated in time. – alortimor Apr 19 '17 at 17:33
  • The chance for all $5$ to fail would be $1/p^5$ where $p$ is the individual failure rate. Yes, in the real world, it could be worse. At the extreme, each scanner could fail with probability $p$ but they could all fail on the same set of viruses. Then it doesn't help to have any more than one. – Ross Millikan Apr 19 '17 at 17:43