Mathematics, 10.09.2019 03:30 MK100

Aspyware is trying to break into a system by guessing its password. it does not give up until it tries 1 million different passwords. what is the probability that it will guess the password and break in if by rules, the password must consist of
(a) 6 different lower-case letters
(b) 6 different letters, some may be upper-case, and it is case-sensitive
(c) any 6 letters, upper- or lower-case, and it is case-sensitive
(d) any 6 characters including letters and digits

Answers: 2

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Answer from: parrazm2022

Probabilities are used to determine the chance of an event.

The probability that it guesses the password for 6 different lower-case letters is 0.006033The probability that it guesses the password for 6 different letters is 0.0000682215The probability that it guesses the password for any 6 letters is 0.00005058012The probability that it guesses the password for any 6 characters is 0.00001760556

(a) 6 different lower-case letters

There are 26 lower-case letters.

So, the number of ways of selecting these letters is:

$\mathbf{n = 26 \times 25 \times 24 \times 23 \times 22 \times 21}$

$\mathbf{n = 165765600}$

The probability is then calculated as:

$\mathbf{Pr =\frac {1000000}{n}}$

So, we have:

$\mathbf{Pr =\frac {1000000}{165765600}}$

$\mathbf{Pr =0.006033}$

The probability that it guesses the password for 6 different lower-case letters is 0.006033

(b) 6 different letters

There are 52 letters (upper and lower cases)

So, the number of ways of selecting these letters is:

$\mathbf{n = 52 \times 51 \times 50 \times 49 \times 48 \times 47}$

$\mathbf{n = 14658134400}$

The probability is then calculated as:

$\mathbf{Pr =\frac {1000000}{n}}$

So, we have:

$\mathbf{Pr =\frac {1000000}{14658134400}}$

$\mathbf{Pr =0.0000682215}$

The probability that it guesses the password for 6 different letters is 0.0000682215

There are 52 letters (upper and lower cases)

Any 6 letters, means that repetition is allowed

So, the number of ways of selecting these letters is:

$\mathbf{n = 52^6}$

$\mathbf{n = 19770609664}$

The probability is then calculated as:

$\mathbf{Pr =\frac {1000000}{n}}$

So, we have:

$\mathbf{Pr =\frac {1000000}{19770609664}}$

$\mathbf{Pr =0.00005058012}$

The probability that it guesses the password for any 6 letters is 0.00005058012

(d) Any 6 characters (letters and digits)

There are 52 letters (upper and lower cases)

There are 10 digits

So, the total character is:

$\mathbf{Total = 52 + 10 = 62}$

Any 6 character, means that repetition is allowed

So, the number of ways of selecting these characters is:

$\mathbf{n = 62^6}$

$\mathbf{n = 56800235584}$

The probability is then calculated as:

$\mathbf{Pr =\frac {1000000}{n}}$

So, we have:

$\mathbf{Pr =\frac {1000000}{56800235584}}$

$\mathbf{Pr =0.00001760556}$

The probability that it guesses the password for any 6 characters is 0.00001760556

Read more about probabilities at:

link

Answer from: jonathon3957

So our limit is 1 million attempts.

a) 6 lower case letters.

Think in 6 blank spaces where you can put one letter, so in each space you have 26 options, so the total of combinations is $26^{6}$ = 308915776.

so if you have 1 million attempts, then 1000000/308915776 = 0.00323

so you have 0.32% chances of breaking the code.

b) 6 different letters, some may be upper-case, and it is case-sensitive

Again 6 blank spaces, in the first you can put 26 letters, and because its case sensitive, you have 52 options, in the second space, you will have 25 posible letters, and 50 options, and so on.

So the total combinations are 52*50*48*46*44*42 = 10608998400

and the probability of breaking it with 1 mill attempts is = 0.009%

same as the first case, but here you have 26 lower case and 26 upper, so you have $52^{6}$ = 19770609664 combinations

and in 1 mill attempts the probability of breaking it is 0.0005%

(d) any 6 characters including letters and digits

Here you have 26 letters and 10 numbers, so the total combinations are :

$36^{6}$ = 2176782336

and the probability of breaking it will be 0.004%

were in all cases you must think that each attempt made by the spyware is a different combination, so the one million tries are different combinations.

Answer from: Quest

Equatorial triangle and obtuce angle

Answer from: Quest

well if he bought them at 5 cedis each sold them at 7 cedis each then he would make a profit of 2 cedis of each book. if in total he made 150 cedis profit that means he sold 75 books. he lost 12 books at the start so that means he originally bought a total of 82 books.

step-by-step explanation: