. Consider Bob, an absent-minded student. Bob has a set of 5 passwords that he u

Question

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Consider Bob, an absent-minded student. Bob has a set of 5 passwords that he uses for all his login needs. He often forgets which password matches with which system so his strategy is to try all of them. Specifically, from his set of passwords he picks one uniformly at random and enters it. If he succeeds all is well; if he fails he removes the password from the set and repeats the process Suppose Bob is currently trying to login into SuperStrictSystem. It has a policy that if a user fails to enter the correct password three consecutive times, the user will be permanently locked out of the system. What is the probability that Bob fails to access the system? Suppose Bob switches to NotSoStrictSystem. This time around, this system will not limit the number of times a user tries to log in. Instead, if a user fails to enter the correct password for k consecutive times, it will delay its response by 2^k seconds. For example, if Bob gets the password right the first time, the delay is just 1 second, etc. Let X(s) be equal to the delay in the response with outcome s. What are the possible values of X(s)? For each such value i, what is P(X = i)? What is E[X]?

Explanation / Answer

The probabilities are much simpler than the ones proposed. Let YY be the number of trials until we get success. Then Y=1,2,3,4,5Y=1,2,3,4,5, each with probability 1/5.

Let X(s) be equal to the delay in response with outcome, s. I got X(s) = {1, 2, 4, 8, 16}

For this, I treated this as a Bernoulli trial, since you can only have a success (correct password) or a failure (wrong password). I'm not sure whether to use the Binomial distribution or just calculate the distribution directly (sum of (X(s)*p(s) until s = 4). Anyways, this is my work:

P(x = 0) = C(5,0)((2/5)^0)(1-2/5)^(5-0) = ((3/5)^5) = (243/3125) P(x = 1) = C(5,1) * ((2/5)^1)((1-2/5)^(5-1)) = 2((3/5)^4) = 162/625 P(x = 2) = C(5,2)*((2/5)^2) * (1-2/5)^(5-2) = 10*((2/5)^2)((3/5)^3) = 216/625 P(x = 3) = C(5,3)((2/5)^3)((1-2/5)^(5-3)) = 10((2/5)^3)((3/5)^2) = 144/625 P(x = 4) = C(5,4)((2/5)^4)((1-2/5)^(5-4)) = 5*((2/5)^4)*(3/5) = 48/625

For this, I used the distribution values from part b as p(s) and the X(s) values from above to get:

1(243/3125)+2(162/625)+4(216/625)+8(144/625)+16(48/625) = 15,783/3125 = E[X]

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