# "A computer system uses passwords [. . .]": Probabilities=?

## Homework Statement

PROBLEM(S):
A computer system uses passwords that contain exactly eight characters, and each character is one of 26 lowercase letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9). Let Ω 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. Suppose that all passwords in Ω are equally likely.

Determine the probability of each of the following:
(a) A
(b) B
(c) A password contains at least 1 integer.
(d) A password contains exactly 2 integers.

ANSWERS:
(a) 0.2448
(b) 4.58E−7
(c) 0.7551
(d) 0.254

## Homework Equations

Multiplication rule.

## The Attempt at a Solution

Could someone please show me how to do (a), so that I can work on the rest on my own? For (a), I tried P(A) = 1/(52^8), but that's wrong.

Mark44
Mentor

## Homework Statement

PROBLEM(S):
A computer system uses passwords that contain exactly eight characters, and each character is one of 26 lowercase letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9). Let Ω 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. Suppose that all passwords in Ω are equally likely.

Determine the probability of each of the following:
(a) A
(b) B
(c) A password contains at least 1 integer.
(d) A password contains exactly 2 integers.

ANSWERS:
(a) 0.2448
(b) 4.58E−7
(c) 0.7551
(d) 0.254

## Homework Equations

Multiplication rule.

## The Attempt at a Solution

Could someone please show me how to do (a), so that I can work on the rest on my own? For (a), I tried P(A) = 1/(52^8), but that's wrong.
For each character chosen, you have 52 possible characters out of a total of 62.

Oh, I see!

So, (a) is (52/62)^8!

Thanks!

I also get (b) to be roughly 4.58E-7, but could you please help me with parts (c) and (d) now?

HallsofIvy
Science Advisor
Homework Helper
Oh, I see!

So, (a) is (52/62)^8!

Thanks!
Assuming that "!" is not a factorial, yes! (And, of course, that fraction can be reduced- 52/62= 26/31.)

For (c), i get [1/10 * 10/62]^8, but that's wrong.

The "!" was not for a factorial; it was just because of my excitement. :)

Could you help me with part (c)?

Dick
Science Advisor
Homework Helper
The "!" was not for a factorial; it was just because of my excitement. :)

Could you help me with part (c)?

The set of all passwords containing at least one integer is equal to the set of ALL passwords minus the set of passwords consisting of only letters. So?

I get 62^8 - 52^8 ~= 1.65E14 passwords containing at least one integer.

Dick
Science Advisor
Homework Helper
I get 62^8 - 52^8 ~= 1.65E14 passwords containing at least one integer.

Now translate that into a probability.