arronjam wrote:

Hi All

I am unable to solve the following question and need help with it.

A Tv show host picks members of the large studio audience at random and asks them what star sign they were born under.

(There are 12 star signs in all and you may assume that the probabilities that a randomly chosen person will be born under each star sign are equal.)

What is the probability that at least two of the first five people picked were born under the same star sign?

I can solve it with the 1-complement approach but i want to solve for individual cases and add them up.

Case 1. Two people born under the same star.

SSDDD S stands for the same star sign and D for a different star sign.

12/12* 1/12* 11/12* 10/12* 9/12 * 5!/3!*2! =9900/12^4

5 stars can be arranged in 5! ways. we need to dived by 3! because the arrangement of three stars has already been counted and divide by 2 to remove the two same star signs eg AA. Is the reasoning correct here?

Similarly Other cases

SSSDD

12/12 *1/12* 1/12*11/12 * 10/12 * 5!/3!*2! = 1100/12^4

SSSSD

12/12 *1/12* 1/12* 1/12 * 11/12 * 5!/4! = 55/12^4

SSSSS

12/12 *1/12* 1/12* 1/12 * 1/12 = 1/12^4

Adding up all these 4 cases yields 11056/12^4 = 0.53 but that is not the correct answer. The correct answer is 0.61 approx. Please help me get the mistake. Thank you

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