Consider the situation of decoding a 6-digit password that consists of the symbols $A$ to $Z$ and $0$ to $9$, where all possible combinations are tried randomly and uniformly.
(a) What is the probability that the correct password will never be entered?
(b) What is the probability that eventually the same combination will be entered two consecutive times?
I am so bad in these combinatorial things.
Can anybody explain me how to calculate these two probabilities, please?
I know that there are $36^5$ possibilities to try.
Edit
For (a) I think I have to look at
$$
\left(\frac{36^5-1}{36^5}\right)^n\to 0\text{ as }n\to\infty.
$$
So the probability that the correct password will never be entered is 0.
Best Answer
Answering (a):
The probability that the correct password will be entered is $\frac{1}{36^6}$
The probability that the correct password will not be entered is $1-\frac{1}{36^6}$
The probability that the correct password will not be entered after $n$ attempts is $(1-\frac{1}{36^6})^n$
Hence the probability that the correct password will never be entered is $\lim\limits_{n\to\infty}(1-\frac{1}{36^6})^n=0$
Answering (b):
Having entered a password:
The probability to enter the same password is $\frac{1}{36^6}$
The probability to enter a different password is $1-\frac{1}{36^6}$
The probability to enter a different password over $n$ attempts is $(1-\frac{1}{36^6})^n$
Hence the probability to always enter a different password is $\lim\limits_{n\to\infty}(1-\frac{1}{36^6})^n=0$
Hence the probability to eventually enter the same password is $1-\lim\limits_{n\to\infty}(1-\frac{1}{36^6})^n=1$