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.

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier

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step-by-step explanation: