Question

A video rental store keeps a list of their top 15 movie rentals each week. This week the list includes 6 action, 4 comedies, 3 dramas, and 2 mysteries. The store manager removes a copy of each of the 15 movies from the shelf, then randomly selects 3 of the 15 to show on the display monitors in the store. What is the probability that she selected 2 comedies and 1 action movie?

Answer 1

To find the probability that the store manager selects 2 comedies and 1 action movie from the list of 15 movies, we can use combinations to calculate the number of favorable outcomes and the total outcomes.

1. **Total Movies and Categories:**
- Action movies (A): 6
- Comedies (C): 4
- Dramas (D): 3
- Mysteries (M): 2
- Total movies = 6 + 4 + 3 + 2 = 15

2. **Total Ways to Choose 3 Movies:**
The total number of ways to choose any 3 movies from 15 is given by the combination formula \( \binom{n}{k} \), where \( n \) is the total number of movies, and \( k \) is the number of movies to choose. Thus, we calculate:
\[
\binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455
\]

3. **Ways to Choose 2 Comedies and 1 Action Movie:**
- The number of ways to choose 2 comedies from 4:
\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]
- The number of ways to choose 1 action movie from 6:
\[
\binom{6}{1} = 6
\]
- So, the total number of ways to choose 2 comedies and 1 action movie is given by multiplying the two results:
\[
6 \times 6 = 36
\]

4. **Calculating the Probability:**
The probability \( P \) that the store manager selected 2 comedies and 1 action movie is given by the ratio of the number of favorable outcomes to the total outcomes:
\[
P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{36}{455}
\]

Thus, the final probability is:

\[
\frac{36}{455}
\]

This fraction cannot be simplified further, so the probability that she selected 2 comedies and 1 action movie is \( \frac{36}{455} \).