Question

A computer 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)

Answer 1

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

$36^6$

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

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

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

$P(A)=P(B)=\dfrac{5}{36}$

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:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$

$exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8$

$ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8$

$oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8$

$uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8$

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

$P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}$

Finally, you can compute the probability of the union using the formula

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Since we already computed all these quantities.

Answer 2

The computer password.

A computer uses a password that can be used for accessing the content of the system. The password is made from the size characters and each is made from a mix of alphabets and 26 letters. The upper and the lower case are not used. Some words and letters denote the events of the password.

Thus the answer is probably is the P(A) = P(B) 5/36

• As per the question, the letter is 26 and 20 integers of 0-9 then upper and case is not used.

• Let B denote the events that are in 0,2,4,6,8 and taking a hacker the password selected at random in order to determine the  25 cases with 4 vacant slots, probability is 25/1296.

Learn more about the computer system.

brainly.com/question/17174600.