Questions: A university class has 22 students: 3 are nursing majors, 9 are psychology majors, and 10 are business majors. (Each student has only one of these majors.) The professor is planning to select two of the students for a demonstration. The first student will be selected at random, and then the second student will be selected at random from the remaining students. What is the probability that the first student selected is a nursing major and the second student is a psychology major? Do not round your intermediate computations. Round your final answer to three decimal places.

Solution Steps

To solve this problem, we need to calculate the probability of two events happening in sequence: first selecting a nursing major and then selecting a psychology major from the remaining students.

1. Calculate the probability of selecting a nursing major first.
2. Calculate the probability of selecting a psychology major from the remaining students after a nursing major has been selected.
3. Multiply these two probabilities to get the final answer.

The probability of selecting a nursing major first is given by the ratio of nursing majors to the total number of students:

\[ P(\text{Nursing First}) = \frac{\text{Number of Nursing Majors}}{\text{Total Students}} = \frac{3}{22} \approx 0.1364 \]

After selecting a nursing major, there are now 21 students remaining. The probability of selecting a psychology major second is given by the ratio of psychology majors to the remaining students:

\[ P(\text{Psychology Second} | \text{Nursing First}) = \frac{\text{Number of Psychology Majors}}{\text{Remaining Students}} = \frac{9}{21} \approx 0.4286 \]

The combined probability of both events occurring (selecting a nursing major first and a psychology major second) is the product of the two individual probabilities:

\[ P(\text{Nursing First and Psychology Second}) = P(\text{Nursing First}) \times P(\text{Psychology Second} | \text{Nursing First}) = \frac{3}{22} \times \frac{9}{21} \approx 0.0584 \]

Final Answer