Probability of a password not having 14 being the first two digits We have a password that consists of $4$ digits, where the password is only numerical values from $0-9$. What is the probability that the first $2$ digits are not $14$?
So, I know that the sample space is $10 \cdot 10 \cdot 10 \cdot 10 = 10 ^4$
but I was told that the total amount of passwords starting with 14 is $10^2$.
Can anyone please explain to me why? I can't seem to wrap around my head on how they got that answer. Is it because it's the first two digits and therefore is
($10$ possible outcomes first digit $\times \ 10$ possible outcomes second digit) = $10^2$
 A: You have a $4$ digit password, and $2$ digits are fixed. The fact that the digits are $1$ and $4$ don't matter, whether the two digits chosen were the same two or two different digits also doesn't matter.

The fact that it is the first two digits that are fixed also doesn't matter.
The important information is that $2$ out of $4$ digits are already chosen, there is no choice regarding these. So when you're "choosing" your four digit password you can ignore the first two digits.
This leaves a choice for $2$ digits, each with $10$ possible states $(0-9)$, so there are $10^2$ possible passwords whose first two digits are $14$.

Since there are $10^2$ passwords whose first digits are $14$, there are $10^4-10^2$ ($10^4$ is the cardinality of the sample space) passwords whose first two digits are not $14$.
So the probability that the first two digits of the password are not $14$ is
$$\frac{10^4-10^2}{10^4}=99\%$$
A: If you are thinking how many cases are there in which the first two numbers is 14(14XX), or any other numbers if the question changed(abXX), you set the first two numbers as known so what you need to consider is only the last two digits(the Xs), and the number of cases for the last two digits is just $10\cdot10=100$, which means the total amount of password starting with "14" is just 100.
