Acomputer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9). uppercase letters are not used. let a denote 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. determine the following probabilities: a. p(a) b. p(b) c. p(a ∩ b) d. p(a ∪ b)

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.

the k side

step-by-step explanation:

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