An antivirus software has a detection rate of 99% for a specific virus and a false positive rate of 0.5%. If a computer is randomly selected and the software indicates a virus, what is the probability that the computer actually has the virus?
Question
An antivirus software has a detection rate of 99% for a specific virus and a false positive rate of 0.5%. If a computer is randomly selected and the software indicates a virus, what is the probability that the computer actually has the virus?
Solution
To solve this problem, we can use Bayes' theorem.
Let's define the events: A: The computer actually has the virus. B: The antivirus software indicates a virus.
We are given the following probabilities: P(A) = 0.99 (detection rate) P(B|¬A) = 0.005 (false positive rate)
We want to find P(A|B), the probability that the computer actually has the virus given that the software indicates a virus.
Using Bayes' theorem, we have: P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) is the probability that the software indicates a virus given that the computer actually has the virus. In this case, it is the detection rate, which is 0.99.
P(B) is the probability that the software indicates a virus. It can be calculated using the law of total probability: P(B) = P(B|A) * P(A) + P(B|¬A) * P(¬A)
P(¬A) is the complement of event A, which is the probability that the computer does not have the virus. It can be calculated as 1 - P(A), which is 1 - 0.99 = 0.01.
Now we can substitute the values into the equation: P(B) = 0.99 * 0.99 + 0.005 * 0.01
Simplifying, we get: P(B) = 0.9801 + 0.00005 P(B) = 0.98015
Finally, we can calculate P(A|B): P(A|B) = (0.99 * 0.99) / 0.98015
Simplifying, we get: P(A|B) ≈ 0.9995
Therefore, the probability that the computer actually has the virus given that the software indicates a virus is approximately 0.9995, or 99.95%.
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