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:
[tex]P(h=c)=\displaystyle\frac{n\ºchoice}{n\ºpossibilities}=\frac{1}{62}[/tex]
A) Because there is no restriction in the first case, the probability than a hacker matchup all the eight characters can be calculated as:
[tex]P(h_{tot}=c_{tot})=\displaystyle \prod_{i=1}^8 P(h_{i}=c_{i})=\frac{1}{62^8}[/tex]
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:
[tex]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}[/tex]
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:
[tex]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}[/tex]
