Just doing some preparation for an exam,
A computer system uses passwords that are $6$ characters and each character is one of the $26$ letters (a-z) or $10$ integers (0-9). Uppercase letters are not used.
Let $A$ be the event that a password begins with a vowel (either a, e, i, o or u) and let $B$ denote the event that a password ends with an even number (either 0, 2, 4, 6 or 8). Suppose a hacker selects a password at random. What are the probabilities $P(A)$, $P(B)$, $P(A \cap B)$ and $P(A \cup B)$ ?
With $P(A)$ the way I've thought about it (though I think it's wrong) is:
Passwords have 6 characters: __ __ __ __ __ __
The total number of passwords is therefore (26 Letters + 10 numbers)$^6$ or $36^6$ by the multiplicative law of probability
36 36 36 36 36 36
But a password which begins with a vowel (5 letters to choose) would be
5 36 36 36 36 36
therefore being $(5*36^5)/36^6 = 5/36$
but that doesn't feel right.. I think what that gives me is the probability of there being at least one vowel, not necessarily being at the start?