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# Probability Question Help

Author Message
Intern
Joined: 01 Jun 2011
Posts: 20

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17 Jul 2018, 23:20
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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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

Math Expert
Joined: 02 Sep 2009
Posts: 52296

### Show Tags

17 Jul 2018, 23:24
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

Kindly re-post accord to the rules. Thank you.
_________________
Re: Probability Question Help &nbs [#permalink] 17 Jul 2018, 23:24
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