Question

christina is randomly choosing three movies to take on vacation from nine action movies, seven science fiction movies, and four comedies. which statement is true? the probability that christina will choose three action movies can be expressed as \\(\frac{_{20}c_3}{_9c_3}\\). the probability that christina will not choose all action movies can be expressed as \\(1 - \frac{_9c_3}{_{20}c_3}\\). the probability that christina will choose three comedies can be expressed as \\(\frac{1}{_4c_3}\\). the probability that christina will not choose all comedies can be expressed as \\(1 - \frac{_4c_3}{_{20}c_3}\\).

Explanation:

Step1: Calculate total number of movies

Total movies = 9 (action) + 7 (sci - fi) + 4 (comedy) = 20. The number of ways to choose 3 movies from 20 is $_{20}C_3$.

Step2: Analyze each option

• Option 1: Probability of choosing three action movies is $\frac{_9C_3}{_{20}C_3}$, so probability of not choosing three action movies should be $1-\frac{_9C_3}{_{20}C_3}$, not $\frac{_{20}C_3}{_9C_3}$, so this is wrong.
• Option 2: Probability of choosing three action movies is $\frac{_9C_3}{_{20}C_3}$, so probability of not choosing three action movies is $1 - \frac{_9C_3}{_{20}C_3}$, this is correct.
• Option 3: Probability of choosing three comedies is $\frac{_4C_3}{_{20}C_3}$, not $\frac{_4C_3}{_4C_3}$ (which is 1), so this is wrong.
• Option 4: Probability of choosing all comedies is $\frac{_4C_3}{_{20}C_3}$, so probability of not choosing all comedies is $1-\frac{_4C_3}{_{20}C_3}$, but the option has $_{20}C_4$ which is wrong (we are choosing 3 movies, so denominator should be $_{20}C_3$), so this is wrong.

Answer:

The second option (The probability that Christina will not choose all action movies can be expressed as $1 - \frac{_9C_3}{_{20}C_3}$)