Respuesta :
Answer:
The probability is [tex]\frac{1}{n}[/tex]
Step-by-step explanation:
If she has n distinct password candidates and only one of which will successfully log her into a secure system, the probability that her first first successful login will be on her k-th try is:
If k=1
[tex]P = \frac{1}{n}[/tex]
Because, in her first try she has n possibles options and just one give her a successful login.
If k=2
[tex]P=\frac{n-1}{n} *\frac{1}{n-1} =\frac{1}{n}[/tex]
Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and 1 of that give her a successful login.
If k=3
[tex]P=\frac{n-1}{n} *\frac{n-2}{n-1} *\frac{1}{n-2} = \frac{1}{n}[/tex]
Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and n-2 that are not correct and after that, she has n-2 possibles options and 1 give her a successful login.
Finally, no matter what is the value of k, the probability that her first successful login will be (exactly) on her k-th try is 1/n
