Answer to Question #125999 in Statistics and Probability for paula cortes

Question

The dance team has 9 dancers to perform different roles in each performance.
The probability that the team selects the same order as last performance is 1 divided by 362880 . A) Explain where each the numerator and denominator come from.
B) What is the probability that the team puts the top hip hop dancer in the third position and top break dancer in the first position? Assume these are not the same people.
C) Create a question, related to the dancers, where you use the indirect method, n(A’), to solve.

Answer 1

A)

There are $9!=362880$ permutations of 9 dancers. So the probability of selecting 1 current permutation is $1/9! = 1/362880$.

B)

There are $7!$ different permutations of dancers where the first dancer and the third dancer are fixed. So the probability of selecting one of such permutations is $7!/9! = 1/72$.

C)

Question

What is the probability that the team puts the top hip hop dancer not in the second position?

Solution

There are $8!$ different permutations of dancers where the second dancer is fixed. So the probability of selecting one of such permutations is $8!/9! = 1/9$. The probability of event where this dancer wasn't placed on the second position is $1 - 1/9 = 8/9$.

Answer: 8/9