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

  • Thread starter s3a
  • Start date
  • #1
s3a
799
8

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.
 

Answers and Replies

  • #2
33,262
4,963

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.
 
  • #3
s3a
799
8
Oh, I see!

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

Thanks!
 
  • #4
s3a
799
8
I also get (b) to be roughly 4.58E-7, but could you please help me with parts (c) and (d) now?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,804
931
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.)
 
  • #6
s3a
799
8
For (c), i get [1/10 * 10/62]^8, but that's wrong.
 
  • #7
s3a
799
8
The "!" was not for a factorial; it was just because of my excitement. :)

Could you help me with part (c)?
 
  • #8
Dick
Science Advisor
Homework Helper
26,258
618
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?
 
  • #9
s3a
799
8
I get 62^8 - 52^8 ~= 1.65E14 passwords containing at least one integer.
 
  • #10
Dick
Science Advisor
Homework Helper
26,258
618
I get 62^8 - 52^8 ~= 1.65E14 passwords containing at least one integer.
Now translate that into a probability.
 

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