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A comedy club hosts an amateur comedy night, a good opportunity for aspiring comedians to perform. Of the $9$ amateur comedians who signed up for tonight's show, $7$ perform farcical comedy. If the host randomly selects $6$ comedians from the sign-up list, what is the probability that all of them practice farcical? Write your answer as a decimal rounded to four decimal places.._________

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Find probabilities using combinations and permutations

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Q. A comedy club hosts an amateur comedy night, a good opportunity for aspiring comedians to perform. Of the

9

amateur comedians who signed up for tonight's show,

7

perform farcical comedy. If the host randomly selects

6

comedians from the sign-up list, what is the probability that all of them practice farcical? Write your answer as a decimal rounded to four decimal places.._________

Calculate Total Combinations: First, calculate the total number of ways to choose $6$ comedians out of $9$ without regard to the type of comedy they perform. This is a combination problem, so we use the formula for combinations: $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items, and $k$ is the number of items to choose. $\newline$ $C(9, 6) = \frac{9!}{6!(9-6)!} = \frac{9!}{6!3!} = \frac{9\times8\times7}{3\times2\times1} = 84$ ways.
Calculate Combinations for Farcical Comedy: Next, calculate the number of ways to choose $6$ comedians from the $7$ who perform farcical comedy. Again, we use the combination formula. $\newline$ $C(7, 6) = \frac{7!}{(6!(7-6)!)} = \frac{7!}{(6!1!)} = \frac{7}{1} = 7$ ways.
Find Probability: Now, find the probability by dividing the number of favorable outcomes by the total number of possible outcomes. $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{7}{84}$ .
Simplify Fraction: Simplify the fraction $\frac{7}{84}$ to get the probability in decimal form. $\newline$ $\frac{7}{84} = \frac{1}{12} \approx 0.0833$ when rounded to four decimal places.

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