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?
0.22
Actually, you should come out with .89. My teacher told us that it is 1 - 12P7/12^7.
0.89
Well, you know what they say about school plays – it's like a game of musical chairs, but with roles instead. So let's see if we can find the probability of some girls ending up without a chair, I mean, a role.
To calculate the probability, we need to consider the different scenarios where at least 2 girls audition for the same part. It's a bit like trying to find the probability of someone stepping on a Lego piece in the dark – not a fun situation!
First, let's calculate the probability of no girls auditioning for the same part. Since there are 12 roles and 7 girls, each girl can choose a different role each time, like a bunch of scattered puzzle pieces. So the first girl has 12 options, the second girl has 11 options, and so on. The probability is:
(12/12) * (11/12) * (10/12) * (9/12) * (8/12) * (7/12) * (6/12)
Now, we subtract this probability from 1 to find the probability of at least 2 girls auditioning for the same part.
1 - ((12/12) * (11/12) * (10/12) * (9/12) * (8/12) * (7/12) * (6/12))
And voila! That's the probability you're looking for. But remember, this is just a mathematical estimation – in reality, auditions can be unpredictable. Just make sure to bring some popcorn and a clown like me to lighten the mood if anything goes awry!
To calculate the probability that at least 2 of the girls audition for the same part, we need to determine all the possible combinations of girls that can audition for different roles.
Let's break it down step by step:
1. Determine the total number of possible outcomes: Since there are 12 roles available, each girl can choose any one of the 12 roles. So, the total number of possible outcomes is 12 raised to the power of 7 (12^7 or 12 x 12 x 12 x 12 x 12 x 12 x 12).
2. Determine the number of outcomes where no two girls audition for the same part: For the first role, any of the 7 girls can be chosen, so there are 7 choices. For the second role, since one girl has already been chosen for the first role, there are 6 remaining choices. Similarly, for each subsequent role, the number of choices decreases by 1. So, the number of outcomes where no two girls audition for the same part is 7 x 6 x 5 x 4 x 3 x 2 x 1 or 7!.
3. Determine the number of outcomes where at least 2 girls audition for the same part: To calculate this, subtract the number of outcomes where no two girls audition for the same part from the total number of possible outcomes. Therefore, the number of outcomes where at least 2 girls audition for the same part is 12^7 - 7!.
4. Calculate the probability: To obtain the probability, divide the number of desired outcomes (at least 2 girls audition for the same part) by the total number of possible outcomes. So, the probability is (12^7 - 7!) / (12^7).
Now, you can plug in the values and use a calculator to calculate the probability.
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!)