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There are twelve prisoners in a certain cell. What is the

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There are twelve prisoners in a certain cell. What is the [#permalink] New post 02 Aug 2003, 23:04
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There are twelve prisoners in a certain cell. What is the probability that:

(1) they all were born in the same month
(2) they all were born in different months
(3) exactly six of them were born in the same month
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 [#permalink] New post 03 Aug 2003, 15:40
P(they all were born in the same month ) = 1 / 12^12 = ???
P(they all were born in different months ) = 12! / 12^12 = ???
P(exactly six of them were born in the same month) =
= 11 * 10 * 9 * 8 * 7 * 6 / 12^12
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 [#permalink] New post 03 Aug 2003, 18:45
The probability that each prisoner was born in January, for example
1/12*1/12*1/12....=1/12^12
If we take all months 12*1/12^12
The probability that all were born in different months
12!/12^12
The probability that exactly six of them were born in the same month
12C6/12^12
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 [#permalink] New post 03 Aug 2003, 21:15
RK73 wrote:
The probability that each prisoner was born in January, for example
1/12*1/12*1/12....=1/12^12
If we take all months 12*1/12^12
The probability that all were born in different months
12!/12^12
The probability that exactly six of them were born in the same month
12C6/12^12


(1) correct
(2) correct
(3) doubt

any 6 are taken (12C6). they were born in the same month (12). other six people have 11*10*9*8*7*6 chances to be born in different months.

Thus, 12C6*12*11*10*9*8*7*6/12^12 correct?
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 [#permalink] New post 04 Aug 2003, 11:25
For Question 1, why would one need to multiply by 12? If one is born
on certain month, say January, every one else should be born on the
same month.

For Question 3, the same logic applies.

What am I missing?
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 [#permalink] New post 12 Jan 2004, 20:03
I got the first 2 right but I think that the 3rd question should be worded as: The probability that exactly six of them were born in the same month
and that the 5 others are all born in different months.
Only then will Stoylar's answer's 12C6*12*11*10*9*8*7*6/12^12 will be good. Because, what if 6 have the same birth months but 2, 3 , 4 or 5 others also have another same birth month? I think worded as I said it, then Stoylar's answer would be right. Any thoughts?
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 [#permalink] New post 15 Jan 2004, 19:19
I think the 1st one should be 12/(12^12) because there are 12 possible months for everyone to be born on the same month. 1/(12^12) implies that there is only one possible way, but everyone can be born in Jan, Feb, Mar...etc all the way to Dec.

For example, if there are only 3 months in a year (Jan , Feb, Mar) and three people, then the total possible combinations are 3/(3^3).

P1 P2 P3
Jan Jan Jan
Feb Feb Feb
Mar Mar Mar
-----------------
Jan Feb Mar
Jan Mar Feb
Feb Jan Mar
Feb Mar Jan
Mar Jan Feb
Mar Feb Jan
21 more ....

I think the last one should be 12C6*12*11*11*11*11*11*11/12^12. Six can be born in the same month and the other six can be born in the same months too...just not the same month as the first six.
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 [#permalink] New post 15 Jan 2004, 21:07
Calnhob, I think your answer should be good. 12C6*12 for 6 having the same birth months, then 11^6 for the other 6 having diff. or same birth months all being accounted for in 11^6. Thus, the final answer, as you said should really be (12C6*11^6) / 12^12. Do you agree that Stoylar's answer represents The probability that exactly six of them were born in the same month and that the 5 others are all born in different months.?
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 [#permalink] New post 15 Jan 2004, 21:59
Paul, the answer Stoylar gave looks good for everyone else having a different month.
  [#permalink] 15 Jan 2004, 21:59
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