Question #175546

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?


1
Expert's answer
2021-03-29T11:09:53-0400

Let A = {router crashes},

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

Then using Bayes formula:

Pr(HnotDoSA)=Pr(AHnotDoS)Pr(HnotDoS)iPr(AHi)Pr(Hi)Pr(H_{notDoS}|A) = \frac{Pr(A|H_{notDoS})Pr(H_{notDoS})}{\sum_iPr(A|H_i)Pr(H_i)}

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

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

Pr(HDoS)=0.2Pr(H_{DoS}) = 0.2

Pr(HnotDoS)=0.8Pr(H_{notDoS}) = 0.8

Pr(HnotDoSA)=0.020.80.020.8+0.950.2=0.078Pr(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%


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