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Five fair dice with faces marked 1,2,3,4,5 and 6 are rolled.

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Five fair dice with faces marked 1,2,3,4,5 and 6 are rolled. [#permalink] New post 08 Nov 2004, 13:03
Five fair dice with faces marked 1,2,3,4,5 and 6 are rolled. Calculate the probability of throwing exactly the same number on at least four dice.

I liked this question quite a lot. Wanted to share it with everyone..
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 [#permalink] New post 08 Nov 2004, 13:17
Denomiator = 6^5 = 7776 possibilities.

Now, for the numerator = total outcomes(which is 6^5) - number of outcomes where all four dice show different numbers (which is 6*5*4*3) -number of outcomes where all show same number (which is 6) = 7776-360 -6 = 7410

Prob is 7410/7776 = 3705/3888. I dont think I am sure of this one. OA pls.
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 [#permalink] New post 08 Nov 2004, 13:19
Well thats wrong..... Shall I give out the OA so fast?
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 [#permalink] New post 08 Nov 2004, 13:20
I know that the numbers are getting unfriednly...some mistake...how about 5 answer choices..
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 [#permalink] New post 08 Nov 2004, 13:21
Also, missed on 'ATleast'...anyway give the answer choices...this is a tough one..
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 [#permalink] New post 08 Nov 2004, 13:22
Wait, not yet. I'm currently working on a report. I'll give an answer within next 15 minutes :lol:
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 [#permalink] New post 08 Nov 2004, 13:27
:-D No answer choices for this one.. sorry venksune.. :)

Its just a Probability sum a like. So I picked it up. IT has a few concepts in it. Lets wait for Paul's Answer.
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 [#permalink] New post 08 Nov 2004, 13:29
First draft with Bernoulli, 6.(5C4.(1/6)^4.(5/6)^1+1/6^5)=26/1296
does not smell so good

Last edited by twixt on 08 Nov 2004, 15:56, edited 3 times in total.
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 [#permalink] New post 08 Nov 2004, 13:35
:arh :fist: In the count of three I need either the answer or Paul's response...can't sleep otherwise peacefully.. :crying
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 [#permalink] New post 08 Nov 2004, 13:39
Too late, damn report! :arh
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 [#permalink] New post 08 Nov 2004, 13:43
The probability of getting four dice to have the same number:

6 / 6^4

So, In the numerator we have: (6/6^4).6= 1/6^2

In the denominator we have 6^5

The result is: (1/6^2)/6^5 = 1/6^7
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 [#permalink] New post 08 Nov 2004, 13:55
Answer should be 25/7776 after the afore mentioned calculations
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 [#permalink] New post 08 Nov 2004, 15:11
The probability of obtaining at least four "sixes" when rolling five dice is as per Paul's solution that is, C(5,4)*5/(6^5). However, there are six marks on each dice {1,2,3,4,5,6} so that ans is

6*C(5,4)*5/(6^5) = 5*5/(6^4) = 25/1296

My humble opinion :?
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 [#permalink] New post 08 Nov 2004, 15:26
Forgot the fact that all five dice could show the same number - In that case ans is
6*[ 5C4*5/(6^5) + 5C5*1/(6^5)] = 13/648 ~ 1/50
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 [#permalink] New post 08 Nov 2004, 15:53
Oxon, you are right, I forgot to consider that there are 6 possible values (1,2,3,4,5,6) that could be same and also when all 5 are same!
Hence, it should be:
[(1/6)^4 * 5/6 * 5C4 + (1/6)^5] * 6 = 26/6^4 = 13/648
You are absolutely right :shock:
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 [#permalink] New post 08 Nov 2004, 16:01
That was a tough one!
ruhi could you please give all 5 answer choices?
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 [#permalink] New post 08 Nov 2004, 16:04
sorry!

OA would do it!
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 [#permalink] New post 08 Nov 2004, 19:57
Extremely good posts (except mine). Well solved.
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 [#permalink] New post 08 Nov 2004, 22:24
Oxorn is RIGHT. Thats the perfect answer. 8-)
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  [#permalink] 08 Nov 2004, 22:24
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Five fair dice with faces marked 1,2,3,4,5 and 6 are rolled.

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