A computer system uses passwords constructed from the 26 letters (a-z) or 10 int

Question

A computer system uses passwords constructed from the 26 letters (a-z) or 10 integers (0-9). Suppose there are 10,000 users on the system with unique passwords. A hacker randomly selects (with replacement) passwords from the potential set. Round your answers to the nearest integer. (a) There are 9, 300 users with unique 6-character passwords on the system and the hacker randomly selects 6-character passwords. What is the mean and standard deviation of the number of attempts before the hacker selects a user password? The mean is, and the standard deviation is. (b) Suppose there are 700 users with unique 3-character passwords on the system and the hacker randomly selects 3-character passwords. What is the mean and standard deviation of the number of attempts before the hacker selects a user password' The mean is and the standard deviation is?

Explanation / Answer

(a) Let X be the number of attempts before the hacker selects a legitimate password.

Then X~Geo(p) where p is the probability of selecting a 6 digit legitimate password.

The pmf of X is given by,

f(x)= (1-p)xp x=0,1,2,.....

= 0 otherwise

Now we need to find the number of all possible 6 digit passwords.

The 1st digit of the password can consist of any one of the 36 ( 26 alphabets+10 integers) character.

Similarly the 2nd,3rd,...,6th digit can be any one of the 36 characters.

Therefore the total number of 6 digit passwords = 366

Hence p = 9300 / 366   (since there are 9300 legit passwords out of 366 passwords)

Required to find,

E(X) = Mean no. of attempts = (1-p)/p = 234062

V(X)= Variance of no. of attempts = (1-p) / p2 = 54785074509

sd(X) = standard deviation of no. of attempts = V(X) = 234062

(b)

Let Y be the number of attempts before the hacker selects a legitimate password.

Then Y~Geo(p) where p is the probability of selecting a 3 digit legitimate password.

The pmf of Y is given by,

f(y)= (1-p)yp y=0,1,2,.....

= 0 otherwise

Now we need to find the number of all possible 3 digit passwords.

The 1st digit of the password can consist of any one of the 36 ( 26 alphabets+10 integers) character.

Similarly the 2nd,3rd digit can be any one of the 36 characters.

Therefore the total number of 6 digit passwords = 363

Hence p = 700 / 363    (since there are 700 legit passwords out of 363 passwords)

Required to find,

E(Y) = Mean no. of attempts = (1-p)/p = 66

V(Y)= Variance of no. of atempts = (1-p) / p2 = 4376

sd(Y) = standard deviation of no. of attempts = V(Y) = 66

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