A college student is taking two courses. The probability she passes the first course is 0.74. The probability she passes the second course is 0.74. The probability she passes at least one of the courses is 0.9324. I would like to know how to approach each of this scenarios, it\'s been so difficult for me to understand the idea of probabilities like. 1. What is the probability she does not pass either course? Give your answer to four decimal places. 2. What is the probability she does not pass both courses? 3. What is the probability she passes exactly one course? 4. Given she passes the first course, what is the probability she passes the second? 5. Given she passes the first course, what is the probability she does not pass the second? Solution P(A)= 0.74 P(B)= 0.74 The probability she passes at least one of the courses = P(AUB) = 0.9324 the probability she passes both courses = P(AINB) = P(A) + P(B) - P(AUB) = 0.74+0.74-0.9324 = 0.5476 1) the probability she does not pass either course = 1 - probability she passes at least one of the courses =1- P(AUB) = 1-0.9324 = 0.0676 2) the probability she does not pass both courses = 1-the probability she passes both courses = 1- P(AINB) = 1-0.5476 = 0.4524 3) the probability she passes exactly one course = probability she passes the first course + probability she passes the second course - 2*(the probability she passes both courses) = 0.74+0.74-2*(0.5476) = 0.3848 4)Given she passes the first course, the probability she passes the second P(B/A)= P(BINA)/P(A) = the probability she passes both courses/probability she passes the first course = 0.5476/0.74 = 0.74 5) Given she passes the first course, the probability she does not pass the second = 1 - P(B/A) = 1-0.74 = 0.26.