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 comedies can be expressed as \\( \frac { 1 } { _ { 4 } c _ { 3 } } \\).
the probability that christina will choose three action movies can be expressed as \\( \frac { _ { 2 0 } c _ { 3 } } { _ { 9 } c _ { 3 } } \\).
the probability that christina will not choose all comedies can be expressed as \\( 1 - \frac { _ { 4 } c _ { 3 } } { _ { 2 0 } c _ { 4 } } \\).
the probability that christina will not choose all action movies can be expressed as \\( 1 - \frac { _ { 9 } c _ { 3 } } { _ { 2 0 } c _ { 3 } } \\).

Explanation:

Step1: Calculate the total number of movies

The total number of movies is \(9 + 7+4=20\) movies. The number of ways to choose \(3\) movies out of \(20\) is given by the combination formula \(_{n}C_{r}=\frac{n!}{r!(n - r)!}\), so \(_{20}C_{3}=\frac{20!}{3!(20 - 3)!}=\frac{20\times19\times18}{3\times2\times1}=1140\)

Step2: Analyze each option

• Option 1:

The number of ways to choose \(3\) comedies out of \(4\) is \(_{4}C_{3}=\frac{4!}{3!(4 - 3)!}=4\). The probability of choosing \(3\) comedies is \(\frac{_{4}C_{3}}{_{20}C_{3}}
eq\frac{1}{_{4}C_{3}}\)

• Option 2:

The number of ways to choose \(3\) action movies out of \(9\) is \(_{9}C_{3}=\frac{9!}{3!(9 - 3)!}=\frac{9\times8\times7}{3\times2\times1}=84\). The probability of choosing \(3\) action movies is \(\frac{_{9}C_{3}}{_{20}C_{3}}
eq\frac{_{20}C_{3}}{_{9}C_{3}}\)

• Option 3:

The probability of choosing all comedies is \(\frac{_{4}C_{3}}{_{20}C_{3}}\), not \(\frac{_{4}C_{3}}{_{20}C_{4}}\). The formula for the probability of not - choosing all comedies is \(P(\text{not all comedies})=1 - P(\text{all comedies})=1-\frac{_{4}C_{3}}{_{20}C_{3}}\), not \(1-\frac{_{4}C_{3}}{_{20}C_{4}}\)

• Option 4:

The probability of choosing all action movies is \(\frac{_{9}C_{3}}{_{20}C_{3}}\). Using the formula \(P(\text{not }A)=1 - P(A)\), the probability of not choosing all action movies is \(1-\frac{_{9}C_{3}}{_{20}C_{3}}\)

Answer:

The probability that Christina will not choose all action movies can be expressed as \(1-\frac{_{9}C_{3}}{_{20}C_{3}}\), so the correct option is the fourth one.