at a party 2/5 of the pizzas ordered had pepperoni. The kids ate only 1/3 of the pepperoni pizza while the parents ate all of th
2 answers:
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First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of
possible passwords.
Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of
possible passwords.
So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is
Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have
As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:
Where x can be any of the 36 characters.
So, we have 25 cases with 4 vacant slots, leading to a probability of
Finally, you can compute the probability of the union using the formula
Since we already computed all these quantities.