A computer system uses password that contain exactly six cha

A computer system uses password that contain exactly six characters and each character is one of the 26 lower-pass letters (a-z) or 26 upper case letters (A-Z) or 10 integers (0-9). Let S denote the set of all possible password, and let A and B denote the event that consist of passwords with only letters or only integers, respectively. Suppose that all password in S are equally likely. Determine the following probabilities 3. P (password contains exactly 2 integers given that it contains at least 1 integer)

Solution

1)
Total number of possible passwords = (26+26+10)^6 = (62)^6
P(A)= (52)^6/(62)^6
P(B)= (10)^6/(62)^6
P(A and B\') = P(Event when password contains only letters and no integers)
=P(A)
Therefore,

P( A/B\') = P( A and B\')/P( B\')
= P(A)/(1-P(B))
       = (52)^6/(62)^6 / (1 - (10)^6/(62)^6)
       = 0.348 Answer
         
2)
P(A\' and B) = P(Event when password contains only integers and no letters)

= P(B) = (10)^6/(62)^6

= 0.0000176 Answer

3)
Let C : Event that password contains exactly two integers.
Let D : Event that password contains atleast one integer.
P(C) = 6C2 * 10^2 * (52)^4
P(D) = 6C1 * 10 * (62)^5
P( C and D) = P ( Event when password contains exactly two integers and must have atleast one integer) = P(C)
P(C/D) = P(C and D)/P(D)
= P(C)/P(D)

= (6C2 * 10^2 * (52)^4)/(6C1 * 10 * (62)^5)

= 0.1995 Answer