Questions: Probabilities of a permutation and a combination There are 9 acts in a talent show. An acrobat, a comedian, a dancer, a guitarist, a juggler, a magician, a singer, a violinist, and a whistler. A talent show host randomly schedules the 9 acts. Compute the probability of each of the following events. Event A: The juggler is first, the comedian is second, and the dancer is third. Event B: The first three acts are the magician, the dancer, and the comedian, in any order. Write your answers as fractions in simplest form. P(A)= P(B)=

Probabilities of a permutation and a combination There are 9 acts in a talent show. An acrobat, a comedian, a dancer, a guitarist, a juggler, a magician, a singer, a violinist, and a whistler. A talent show host randomly schedules the 9 acts. Compute the probability of each of the following events. Event A: The juggler is first, the comedian is second, and the dancer is third. Event B: The first three acts are the magician, the dancer, and the comedian, in any order. Write your answers as fractions in simplest form. P(A)= P(B)=
Transcript text: Probabilities of a permutation and a combination There are 9 acts in a talent show. An acrobat, a comedian, a dancer, a guitarist, a juggler, a magician, a singer, a violinist, and a whistler. A talent show host randomly schedules the 9 acts. Compute the probability of each of the following events. Event A: The juggler is first, the comedian is second, and the dancer is third. Event B: The first three acts are the magician, the dancer, and the comedian, in any order. Write your answers as fractions in simplest form. \[ \begin{array}{l} P(A)=\square \\ P(B)=\square \end{array} \]
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Solution

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Solution Steps

To solve the given problem, we need to calculate the probabilities of specific events in a permutation of 9 acts.

Event A: The juggler is first, the comedian is second, and the dancer is third.
  1. There is only one way to arrange the juggler, comedian, and dancer in the first three positions.
  2. The remaining 6 acts can be arranged in any order.
  3. The total number of permutations of 9 acts is 9!.
Event B: The first three acts are the magician, the dancer, and the comedian, in any order.
  1. There are 3! ways to arrange the magician, dancer, and comedian in the first three positions.
  2. The remaining 6 acts can be arranged in any order.
  3. The total number of permutations of 9 acts is 9!.
Step 1: Calculate Total Number of Permutations

The total number of permutations of 9 acts is given by \(9!\): \[ 9! = 362880 \]

Step 2: Calculate Permutations for Event A

For Event A, the juggler is first, the comedian is second, and the dancer is third. The remaining 6 acts can be arranged in any order: \[ 6! = 720 \] The probability of Event A is: \[ P(A) = \frac{6!}{9!} = \frac{720}{362880} = \frac{1}{504} \]

Step 3: Calculate Permutations for Event B

For Event B, the first three acts are the magician, the dancer, and the comedian, in any order. There are \(3!\) ways to arrange these three acts, and the remaining 6 acts can be arranged in any order: \[ 3! \times 6! = 6 \times 720 = 4320 \] The probability of Event B is: \[ P(B) = \frac{3! \times 6!}{9!} = \frac{4320}{362880} = \frac{1}{84} \]

Final Answer

\[ \boxed{P(A) = \frac{1}{504}} \] \[ \boxed{P(B) = \frac{1}{84}} \]

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