Answer :
To determine which statement is true, we need to evaluate the probabilities given in each statement step-by-step.
First, let's recall the combination formula, which is used to determine the number of ways to choose [tex]\( r \)[/tex] items from [tex]\( n \)[/tex] items without regard to the order:
[tex]\[ {}_nC_r = \frac{n!}{r! (n - r)!} \][/tex]
Given data:
- There are 9 action movies, 7 science fiction movies, and 4 comedies.
- Christina is choosing 3 movies out of the total 20 movies.
### Statement 1:
"The probability that Christina will choose three comedies can be expressed as [tex]\(\frac{1}{{}_4C_3}\)[/tex]"
Let's compute this probability step-by-step.
The number of ways to choose 3 comedies out of 4 is:
[tex]\[ {}_4C_3 = \frac{4!}{3!(4 - 3)!} = \frac{4}{1} = 4 \][/tex]
Therefore, the probability that Christina will choose exactly 3 comedies out of the 4 available is:
[tex]\[ \frac{1}{4} = 0.25 \][/tex]
Thus, the statement is true, and the probability is 0.25.
### Statement 2:
"The probability that Christina will choose three action movies can be expressed as [tex]\(\frac{{}_20C_3}{{}_3C_3}\)[/tex]"
Here, we need to calculate the number of ways Christina can choose any 3 movies from the total 20 movies and from 3 specific action movies.
The number of ways to choose 3 movies out of 20 movies is:
[tex]\[ {}_20C_3 = \frac{20!}{3!(20 - 3)!} = 1140 \][/tex]
The number of ways to choose 3 out of the 3 specified action movies is:
[tex]\[ {}_3C_3 = 1 \][/tex]
Therefore, the fraction:
[tex]\[ \frac{{}_20C_3}{{}_3C_3} = \frac{1140}{1} = 1140 \][/tex]
Thus, the statement is true and equals 285.0.
### Statement 3:
"The probability that Christina will not choose all comedies can be expressed as [tex]\(1-\frac{{}_4C_3}{{}_20C_4}\)[/tex]"
Let's compute this probability.
The number of ways to choose 3 comedies out of 4 is:
[tex]\[ {}_4C_3 = 4 \][/tex]
The number of ways to choose any 4 movies out of 20 movies is:
[tex]\[ {}_20C_4 = 4845 \][/tex]
The probability of choosing all comedies:
[tex]\[ \frac{4}{4845} \][/tex]
The probability that Christina will not choose all comedies is:
[tex]\[ 1 - \frac{4}{4845} \approx 1 - 0.0008256 = 0.9991744 \][/tex]
Thus, the statement is true, and the probability is approximately 0.9991744066047472.
### Statement 4:
"The probability that Christina will not choose all action movies can be expressed as [tex]\(1-\frac{C_3}{3C_3}\)[/tex]"
To compute this, first note that [tex]\({}_9C_0\)[/tex] should be equal to 1 for the first term of the fraction to be correctly computed.
The number of ways to choose 3 action movies out of 9 is:
[tex]\[ {}_9C_3 = 84 \][/tex]
However, the probability of the form expressed as [tex]\(\frac{C_3}{3C_3}\)[/tex] isn't accurately representing the intention from the context presented.
Therefore, the probability statement simplifies to understanding that given total choosing subsets.
Thus, the referenced miss-expression leads to negative interpretations:
Thus, the last statement is looked at separately, misses true conditional where calculation errors are less likely as -83.0.
In summary:
- The first statement is true.
- The second statement matches the data's computational logic.
- The third statement probabilistically holds.
- The fourth statement deems significant as inaccurate computation steps shown erroneous context.
First, let's recall the combination formula, which is used to determine the number of ways to choose [tex]\( r \)[/tex] items from [tex]\( n \)[/tex] items without regard to the order:
[tex]\[ {}_nC_r = \frac{n!}{r! (n - r)!} \][/tex]
Given data:
- There are 9 action movies, 7 science fiction movies, and 4 comedies.
- Christina is choosing 3 movies out of the total 20 movies.
### Statement 1:
"The probability that Christina will choose three comedies can be expressed as [tex]\(\frac{1}{{}_4C_3}\)[/tex]"
Let's compute this probability step-by-step.
The number of ways to choose 3 comedies out of 4 is:
[tex]\[ {}_4C_3 = \frac{4!}{3!(4 - 3)!} = \frac{4}{1} = 4 \][/tex]
Therefore, the probability that Christina will choose exactly 3 comedies out of the 4 available is:
[tex]\[ \frac{1}{4} = 0.25 \][/tex]
Thus, the statement is true, and the probability is 0.25.
### Statement 2:
"The probability that Christina will choose three action movies can be expressed as [tex]\(\frac{{}_20C_3}{{}_3C_3}\)[/tex]"
Here, we need to calculate the number of ways Christina can choose any 3 movies from the total 20 movies and from 3 specific action movies.
The number of ways to choose 3 movies out of 20 movies is:
[tex]\[ {}_20C_3 = \frac{20!}{3!(20 - 3)!} = 1140 \][/tex]
The number of ways to choose 3 out of the 3 specified action movies is:
[tex]\[ {}_3C_3 = 1 \][/tex]
Therefore, the fraction:
[tex]\[ \frac{{}_20C_3}{{}_3C_3} = \frac{1140}{1} = 1140 \][/tex]
Thus, the statement is true and equals 285.0.
### Statement 3:
"The probability that Christina will not choose all comedies can be expressed as [tex]\(1-\frac{{}_4C_3}{{}_20C_4}\)[/tex]"
Let's compute this probability.
The number of ways to choose 3 comedies out of 4 is:
[tex]\[ {}_4C_3 = 4 \][/tex]
The number of ways to choose any 4 movies out of 20 movies is:
[tex]\[ {}_20C_4 = 4845 \][/tex]
The probability of choosing all comedies:
[tex]\[ \frac{4}{4845} \][/tex]
The probability that Christina will not choose all comedies is:
[tex]\[ 1 - \frac{4}{4845} \approx 1 - 0.0008256 = 0.9991744 \][/tex]
Thus, the statement is true, and the probability is approximately 0.9991744066047472.
### Statement 4:
"The probability that Christina will not choose all action movies can be expressed as [tex]\(1-\frac{C_3}{3C_3}\)[/tex]"
To compute this, first note that [tex]\({}_9C_0\)[/tex] should be equal to 1 for the first term of the fraction to be correctly computed.
The number of ways to choose 3 action movies out of 9 is:
[tex]\[ {}_9C_3 = 84 \][/tex]
However, the probability of the form expressed as [tex]\(\frac{C_3}{3C_3}\)[/tex] isn't accurately representing the intention from the context presented.
Therefore, the probability statement simplifies to understanding that given total choosing subsets.
Thus, the referenced miss-expression leads to negative interpretations:
Thus, the last statement is looked at separately, misses true conditional where calculation errors are less likely as -83.0.
In summary:
- The first statement is true.
- The second statement matches the data's computational logic.
- The third statement probabilistically holds.
- The fourth statement deems significant as inaccurate computation steps shown erroneous context.
