Question

At the beginning of the semester, a professor tells students that if they study for the tests, then there is a 55% chance they will get a B or higher on the tests. If they do not study, there is a 20% chance that they will get a B or higher on the tests. The professor knows from prior surveys that 60% of students study for the tests. The probabilities are displayed in the tree diagram. A tree diagram. Random student to studies for test is 0.6, and to does not study for test is 0.4. Studies for test to Gets B or higher is 0.55; does not get B or higher is 0.45. Does not study for test to Gets B or higher is 0.20; does not get B or higher is 0.80. The professor informs the class that there will be a test next week. What is the probability that a randomly selected student studied for the test if they pass it with a B or higher? 0.20 0.55 0.60 0.80

Answer 1

The probability that a randomly selected student studied for the test if they pass it with a B or higher is approximately 80.49%.

Step-by-step explanation:

The question is asking for the probability that a student has studied for a test given that they got a B or higher on it. This is a question of conditional probability, which requires the use of Bayes' Theorem or a direct calculation through a tree diagram.

To calculate this probability, we consider the following:

• Probability of studying and getting B or higher = P(Study) × P(B or higher | Study) = 0.6 × 0.55
• Probability of not studying and getting B or higher = P(Not Study) × P(B or higher | Not Study) = 0.4 × 0.20

Therefore, the probability that a student studied given they got a B or higher is:

P(Study | B or higher) = P(Study and B or higher) / P(B or higher)

P(B or higher) = P(Study and B or higher) + P(Not Study and B or higher)

P(Study and B or higher) = 0.6 × 0.55 = 0.33

P(Not Study and B or higher) = 0.4 × 0.20 = 0.08

P(B or higher) = 0.33 + 0.08 = 0.41

So the conditional probability is:

P(Study | B or higher) = 0.33 / 0.41 ≈ 0.8049 or 80.49%