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There are 26 lowercase and 26 capital letters as well as 10 numerical digits, which makes for 62 total possible characters per spot. Since each of the 13 spots can be any of these 62 digits, the total number of possibilities will be 62^13. Because the hacker has to guess on the first try, he cannot rule out any possibilities. The probability that he will guess the password on the first try is 1/(62^13).
The probability that the hacker guesses the password on his first try is 1/(62^(13) )
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
What is the rule of product in combinatorics?
If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in [tex]p \times q[/tex] ways.
Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.
Thus, doing A then B is considered same as doing B then A
Assuming the hacker will try all alphanumeric characters randomly, each of those 13 characters can have (26+26+10= 62) possible characters (26 from lowercase, 26 from uppercase and 10 from digits).
Each of those 13 character places in password is independent of each other, and can have any one of the 62 characters available.
Thus, by the rule of product, the number of possible passwords are:
[tex]62 \times 62 \times \cdots \times 62 \text{ (13 times )} = 62^{13}[/tex]
Let E = event that hacker cracks the password in a random trial
Then, there are [tex]62^{13}[/tex] total passwords he can make, and for event E to be true, he can make only 1 pasword.
Thus, we get: number of ways in which E can be done = 1
number of distinct outcomes possible = [tex]62^{13}[/tex]
Thus, we get:
[tex]P(E) = \dfrac{1}{62^{13}} \approx 4.99 \times 10^{-24}[/tex] (very very small probability)
Thus, the probability that the hacker guesses the password on his first try is 1/(62^(13) )
Learn more about probability here:
brainly.com/question/1210781
