Answer to Question #175546 in Statistics and Probability for santhi

Question

A router subjected to denial-of-service attack (DoS) will crash with probability of 95%. The router can also crash for other reasons with probability of 2%. DoS attacks occur with probability of 0.2. If the router crashes, what is the probability that the cause is not DoS?

Answer 1

Let A = {router crashes},

$H_i = \{attack\ i \ occur\} \forall i \in \{DoS, not DoS\}$

Then using Bayes formula:

$Pr(H_{notDoS}|A) = \frac{Pr(A|H_{notDoS})Pr(H_{notDoS})}{\sum_iPr(A|H_i)Pr(H_i)}$

$Pr(A|H_{notDoS}) = 0.02$

$Pr(A|H_{DoS}) = 0.95$

$Pr(H_{DoS}) = 0.2$

$Pr(H_{notDoS}) = 0.8$

$Pr(H_{notDoS}|A) = \frac{0.02 * 0.8}{0.02 * 0.8 + 0.95 * 0.2} = 0.078$

So, required probability is approximately 7.8%