Question

A computer system uses passwords that are exactly 8 characters and each character is one of the 26 lowercase letters (a-z), uppercase letters (A-Z), or 10 integers (0-9). Let A denote the subset of possible passwords that do not begin with a vowel (a, e, i, o, u, A, E, I, O, U) and B denote the subset of possible passwords that do not end with a number (0-9).(a) Suppose the hacker selects a password at random. what is theprobability that your password is selected?(b) Suppose a hacker knows your password is in event A and selectsa password at random from this subset. what is the probability thatyour password is selected.(c) Suppose a hacker knows your password is in A and B and selectsa password at random from this subset. what is the probability thatyour password is selected?

Answer 1

Answer:

A) the probability than your password is selected is 4.58·10^(-15)

B) If your password is in event A, the probability than your password is selected is 5.46·10^(-15)

C) If your password is in event A and B, the probability than your password is selected is 6.51·10^(-15)

Step-by-step explanation:

The probability than one of the characters c in one specific position is equal to the selected by the hacker h is:

$P(h=c)=\displaystyle\frac{n\ºchoice}{n\ºpossibilities}=\frac{1}{62}$

A) Because there is no restriction in the first case, the probability than a hacker matchup all the eight characters can be calculated as:

$P(h_{tot}=c_{tot})=\displaystyle \prod_{i=1}^8 P(h_{i}=c_{i})=\frac{1}{62^8}$

B) If your password is in event A, the first character has limited its number of possibilities to 52, while the other 7 characters keeps the probability of 1/62 each. Therefore:

$P(h_1\not\in A/ h_{2-8}=c_{2-8})=\displaystyle P(h_{1}=c_{1} \not\in A)\cdot\prod_{i=1}^7 P(h_{i}=c_{i})=\frac{1}{52}\frac{1}{62^7}$

C) If your password is in event A and B, the first and last character has limited its number of possibilities to 52, while the other 6 characters keeps the probability of 1/62 each. Therefore:

$P(h_1\not\in (A\cap B)/ h_{2-7}=c_{2-7})=\displaystyle P(h_{1,8}=c_{1,8} \not\in (A\cap B))\cdot\prod_{i=1}^7 P(h_{i}=c_{i})=\frac{1}{52^2}\frac{1}{62^6}$