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aalyn [17]
3 years ago
6

A computer password consists of eight characters. How many different passwords are possible if each character may be any lowerca

se letter or digit? How many different passwords are possible if each character may be any lowercase letter or digit, and at least one character must be a digit? A computer system requires that passwords contain at least one digit. If eight characters are generated at random, and each is equally likely to be any of the 26 letters or 10 digits, what is the probability that a valid password will be generated?
Mathematics
1 answer:
tankabanditka [31]3 years ago
4 0

Answer:

1. 2,821,109,907,456 passwords

2. 2,612,282,842,880 passwords

3. 0.93

Step-by-step explanation:

Number of lower case letters = 26

Number of digits = 10

Total Characters = 26 + 10 = 36

Length of password = 8

1. Each character can be chosen from 36 objects (26 lowercase letters and 10 digits).

Since there's no restriction on repetition of characters, there are

(26+10)^8 different passwords

= 36^8

= 2,821,109,907,456 passwords

2.

For atleast one character to be a must be a digit means 1 character or 2 character or 3 characters... and so on

So we calculate by looking at no character being a digit

and then subtract the result from the total(36^8).

No character being a digit = All the characters are only letters

I.e. 26^8

Atleast one character being a digit = 36^8 - 26^8

= 2,821,109,907,456 - 208,827,064,576

= 2,612,282,842,880 passwords

3.

Probability = Number of possible outcomes/Number of Total outcomes

Number of possible outcomes = Number of Atleast 1 digit = 36^8 - 26^8 (as in question 2)

Total outcomes = 36^8

Probability = (36^8 - 26^8) / 36^8

= 2,612,282,842,880 / 2,821,109,907,456

= 0.92597698373108952

= 0.93 (approximated)

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