Cyber-attacks on a government website occur according to a Poisson process with

Question

Cyber-attacks on a government website occur according to a Poisson process with an average of 5 minutes between each attack. (a) If no cyber-attacks have occurred between 9:00pm and 9:04pm, what is the probability that at least one cyber-attack will occur before 9:05pm? (b) If no cyber-attacks occur between 9:00pm and 9:05pm, what is the expected waiting time until the next cyber-attack? (c) The security team needs to take down the website's firewall for 30 seconds for an important upgrade. They wait until 9:10pm, and after still not observing any cyber-attacks decide to proceed with the upgrade. What is the probability that any cyber-attacks will occur while the firewall is down? (d) After installing the update, the security team discovers at 9:10:30pm that one cyber-attack did occur at some time while the firewall was down. What is the probability that this attack occurred in the last 10 seconds, that is, between 9:10:20pm and 9:10:30pm? (e) What is the standard deviation for the random variable that describes the number of seconds into the upgrade until the attack occurred in (d)? (f) Thankfully, the security team is able to deal with the threat of one cyber-attack while the firewall is down. What is the probability that more than 1 cyber-attack could have occurred during the upgrade? (g) What is the probability that at least 10 cyber-attacks will occur in any given hour?

Explanation / Answer

Solution: "As per chegg policy, I can give the solution of 4 subparts only. Here I am giving solution of 5 subparts for your convinience."

(a)

P(X> =1) gives the probability of atleast one attack before 9:05 PM

        = 1 - P(0), that is 1 - prob of no attacks

For the 1 minute period between 9:04 and 9:05, arrival rate = 1/5

Reqd prob p = 1 -e^-.2 = 0.1813

(b)

The arrival (attacks) are independent of one another. So, expected waiting time till next attack = 5 min

(c)

We require X= Po(30 sec)

                  = 1 - e^.1 = 0.0952

(d)

X= Po(10 sec) = e-1/30 = 0.967

(g)

Atleast 10 cyber attacks in any given hours = P(X>=10)

   = P(10) + P(11) + P(12) with arrival rate, = 12

   = 0.10484 + 0.11437 + 0.11437 = 0.33358 0.334

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