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?

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It's perfectly right. The probability that there is at least one vowel (not necessarily at the start) is not as simple as this. –  Florian Oct 25 '11 at 6:21
The counting approach you used is fine. However, it can be done more simply. The probability that the first symbol of the password is simply $5/36$, since there are $36$ possibilities for the first symbol, all equally likely, and only $5$ of them "work." Finding $P(B)$ is also straightforward. Finding $P(A \cap B)$ is harder. There your counting approach will work nicely. So will conditional probability, if you have been introduced to it. –  André Nicolas Oct 25 '11 at 6:27

The claim that "your answer is indeed the probability that there is one vowel" is wrong, regardless of how one interprets it: the probability that there is at least one vowel is $1-\left(\frac{31}{36}\right)^6 \approx 0.5923$, while the probability that there is exactly one vowel is $6\cdot\left(\frac{5}{36}\right)\cdot\left(\frac{31}{36}\right)^5 \approx 0.3946$. Neither of those equals $\frac{5}{36}$. –  Ilmari Karonen Oct 25 '11 at 9:36