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What is the approx. probability that at least two of the first five pe

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What is the approx. probability that at least two of the first five pe  [#permalink]

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New post Updated on: 18 Jul 2018, 23:29
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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 approx. probability that at least two of the first five people picked were born under the same star sign?

A 0.60
B 0.50
C 0.62
D 0.38
E 0.55

Originally posted by arronjam on 17 Jul 2018, 23:56.
Last edited by arronjam on 18 Jul 2018, 23:29, edited 1 time in total.
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Re: What is the approx. probability that at least two of the first five pe  [#permalink]

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New post 18 Jul 2018, 02:29
Total Probability - No one born under the same sign
1 - none born
1- (12/12) *(11/12)*(10/12)*(9/12)*(8/12)
1- 0.38
0.62
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Re: What is the approx. probability that at least two of the first five pe  [#permalink]

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New post 18 Jul 2018, 02:44
learnwithrohan wrote:
Total Probability - No one born under the same sign
1 - none born
1- (12/12) *(11/12)*(10/12)*(9/12)*(8/12)
1- 0.38
0.62



Hi can you comment on the solution that i provided though that solution does not give the correct answer?
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Re: What is the approx. probability that at least two of the first five pe  [#permalink]

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New post 18 Jul 2018, 02:46
If every person picked up has a different start sign :

then 5 people could be picked up with probability : 12*11*10*9*8/12*12*12*12*12 { after picking one start sign your pool of star signs reduces by 1).
= 0.38

Now if all people do not have different star signs then at least 2 of them must have same star sign.
Thus probability that at least 2 people have same star signs = 1 - (probability that all have different star signs)
= 1-0.38 = 0.62

Option C
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Re: What is the approx. probability that at least two of the first five pe  [#permalink]

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New post 18 Jul 2018, 03:24
arronjam wrote:
learnwithrohan wrote:
Total Probability - No one born under the same sign
1 - none born
1- (12/12) *(11/12)*(10/12)*(9/12)*(8/12)
1- 0.38
0.62



Hi can you comment on the solution that i provided though that solution does not give the correct answer?


Hey in the first STEP
SSDDD it is actually SSDEF , what I mean is all the other signs are different hence you can not take all of them as D
therefore , when you multiply by the final permutation is should be 5!/2! [as two SS are same ]
similiarly in second step it is 5!/3! only

Hope it helps :)
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Re: What is the approx. probability that at least two of the first five pe  [#permalink]

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New post 18 Jul 2018, 23:17
learnwithrohan wrote:
arronjam wrote:
learnwithrohan wrote:
Total Probability - No one born under the same sign
1 - none born
1- (12/12) *(11/12)*(10/12)*(9/12)*(8/12)
1- 0.38
0.62



Hi can you comment on the solution that i provided though that solution does not give the correct answer?


Hey in the first STEP
SSDDD it is actually SSDEF , what I mean is all the other signs are different hence you can not take all of them as D
therefore , when you multiply by the final permutation is should be 5!/2! [as two SS are same ]
similiarly in second step it is 5!/3! only

Hope it helps :)



I think when we multiply by 5! we are double counting the permutations so it must be 5/3!2! 3 for removing the double counting and 2 for the same star signs SS?
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Re: What is the approx. probability that at least two of the first five pe  [#permalink]

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New post 18 Jul 2018, 23:34
Hi Bunuel

Can you help me get what is wrong in the following solution?

Hi All

Can someone help me with the following question?

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. Please help me get the mistake. Thank you
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Re: What is the approx. probability that at least two of the first five pe  [#permalink]

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New post 19 Jul 2018, 00:43
@aaronjam
hey Sorry, I misread what you had done.
So far you are absolutely correct, but you are missing out on two conditions
Case 1(when two people are born under the same sign and the remaining three people are also born under the same sign, but different from the first sign i.e Jan Jan Feb Feb Feb)

Case 2 (when two people are born under the same sign, two out of the remaining three are born under the same sign, but different from the first sign, and the last is born under a different sign from these four i.e Jan Jan Feb Feb Mar)

Once you add these two cases you will get the correct solution.
Hope this helps :)
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Re: What is the approx. probability that at least two of the first five pe &nbs [#permalink] 19 Jul 2018, 00:43
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