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.
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Total Movies and Categories:
- Action movies (A): 6
- Comedies (C): 4
- Dramas (D): 3
- Mysteries (M): 2
- Total movies = 6 + 4 + 3 + 2 = 15
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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: \[ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \]
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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: \[ 6 \times 6 = 36 \]
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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 \( \frac{36}{455} \).