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7 girls audition for 12 roles in a school play. What is the probability that at least 2 of the girls audition for the same part?

• algebra 2 help please -

This is one minus the probability that all the girls audition for different roles. The total number of ways of assigning roles to the girls is 12^7, because to each of the 7 girls, you have a choice of 12 roles.

Then if each girl is to receive a different role, then there are 12!/5! possibilities for that. If you start assigning roles to the girls, then for the first girl, there are 12 choices, but for the next you have to choose one of the 11 different ones, so 11 for the next, and then one of the 10 remaining for the next etc. etc., and this is 12*11*10*...*6 = 12!/(12-7)! =12!/5!

The probability that a random assignment of one of the 12^7 roles would happen to be one of the 12!/5! roles where each girl has a different role, is

(12!/5!)/12^7 = 12!/(12^7 5!)

Then the probability that two or more girls addition for the same part is the probability that not all the girls are assigned different roles, this is thus:

1 - 12!/(12^7 5!)

• algebra 2 help please -

0.22

• algebra 2 help please -

Actually, you should come out with .89. My teacher told us that it is 1 - 12P7/12^7.

• algebra 2 help please -

HELP!!!!

• algebra 2 help please -

0.89

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