2. A computer system uses passwords that contain exactly six characters, and eac

Question

2. A computer system uses passwords that contain exactly six characters, and each character is 1 of the 26 lowercase letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9). Let denote the set of all possible passwords, and let A and B denote the events that consist of passwords with only letters or only integers, respectively. Determine the number of passwords in each of the following events.

(a)

(b) A

(c) A B

(d) Passwords that contain at least 1 integer

(e) Passwords that contain exactly 1 integer Determine the probability for each of the following:

(f) Password contains all lowercase letters given that it contains only letters

(g) Password contains at least 1 uppercase letter given that it contains only letters

(h) Password contains only even numbers given that it contains all numbers

Explanation / Answer

Although not explicitly stated in the question, it is implied that the selection is with replacement meaning that the same character can appear any number of times in the password.

Back-up Theory

Number of ways of selecting r things out of n things is given by nCr= (n!)/{(r!)(n - r)!}…(1)

Number of ways of selecting r things out of n things when the same thing can be selected any number of times (i.e., with replacement) is given by nr…………………………………..(2)

Part (a)

Number of passwords in ?

In total there are (26 + 26 + 10) = 62 characters available out of which the password can have only 8. So, vide (2), 628 passwords are possible. ANSWER 1

Part (b)

Number of passwords in A

In total there are (26 + 26) = 52 characters (exclusive letters) available out of which the password can have only 8. So, vide (2),528 passwords are possible. ANSWER 2

Part (c)

Number of passwords in (AC ? BC)

By De Morgan’s Law, (AC ? BC) = (A ? B)C

Number of passwords in B

In total there are 10 characters (exclusive integers) available out of which the password can have only 8. So, vide (2), 108 passwords are possible.

Noting that A and B are disjoint sets,

Number of passwords in (A ? B) = Number of passwords in A + Number of passwords in B

= 528 + 108

Hence, Number of passwords in (AC ? BC), i.e., (A ? B)C

= Number of passwords in ? - Number of passwords in (A ? B)

= 628 – (528 + 108) ANSWER

Part (d)

Number of passwords with at least one integer

= Total number of passwords - Number of passwords with no integer

= Number of passwords in ? - Number of passwords in A

= 628 – 528ANSWER

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