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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, 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. Round your answers to four decimal places (e.g. 98.7654).

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MrRoyal

Question continuation

Determine the following probabilities:

a. P(A)

b. P(B)

c. P(A ∩ B)

d. P(A ∪ B)

Answer:

a. P(A) = 0.1389

b. P(B) = 0.1389

c. P(AnB) = 0.0193

d. P(AuB) = 0.2585

Explanation:

Given

Password length = 6

Letters (a-z) = 26

Integers (0-9) = 10

Total usable characters = 26 + 10 = 36

a. P(A) = Probability that a password begins with vowel (a,e,i,o,u)

Probability = Number of required outcomes/ Number of possible outcomes

Number of required outcomes = Number of vowels = 5

Number of possible outcomes = Total usable characters = 36

P(A) = 5/36

P(A) = 0.13888888888

P(A) = 0.1389

b. P(B) = Probability that the password ends with an even number (0,2,4,6,8)

Probability = Number of required outcomes/ Number of possible outcomes

Number of required outcomes = Number of even numbers = 5

Number of possible outcomes = Total usable characters = 36

P(B) = 5/36

P(B) = 0.13888888888

P(B) = 0.1389

c. P(AnB)

This means that the probability that a password starts with a vowel and ends with an even number

P(AnB) = P(A) and P(B)

P(AnB) = P(A) * P(B)

P(AnB) = 5/36 * 5/36

P(AnB) = 25/1296

P(AnB) = 0.01929012345

P(AnB) = 0.0193 ----_---- Approximately

d. P(AuB)

This means that the probability that a password either starts with a vowel or ends with an even number

P(AuB) = P(A) or P(B)

P(AuB) = P(A) + P(B) - P(AnB)

P(AuB) = 5/36 + 5/36 - 25/1296

P(AuB) = 335/1296

P(AuB) = 0.25848765432

P(AuB) = 0.2585 ----_---- Approximately