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# 6 fair dice are tossed. What is the probability that at

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Director
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6 fair dice are tossed. What is the probability that at [#permalink]

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09 Oct 2007, 12:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

6 fair dice are tossed. What is the probability that at least two of them show the same face.

Pleas provide your explanations, thank you.

original post available at:

http://www.gmatclub.com/forum/t53592

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SVP
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09 Oct 2007, 13:02
To me, we should consider the opposite case : all dices have different numbers.

P(At least 2) = 1 - P(all different)
= 1 - (6*5*4*3*2*1) / (6^6)
= 1 - 20/6^4

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Senior Manager
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Location: Greater New York City area
Schools: Tuck, Ross (R1), Duke, Tepper, ISB (R2), Kenan Flagler (R2)

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09 Oct 2007, 13:16
Atleast 2 are dice need to have same number, which means 2,3,4,5 and 6 can have same.

So we can compute the prob of all have different and subtract that from 1 which will give us the answer.

For the first dice, the prob is 1 since any value is fine.
For the second, the prob has to be other than first. So 5/6
For third, the prob has to be other than first and second. So 4/6.
For fourth, 3/6
For fifth. 2/6
For sixth, only one remains so 1/6.

So the probability is (1).(5/6).(4/6).(3/6).(2/6).(1/6) = 5/324.
So the answer is 1-5/324

Answer = 319/324

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Director
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Re: PERMUTATION & COMBINATION gurus!!! [#permalink]

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09 Oct 2007, 13:55
Thank you for your explanations, clear now.

one question...if it were not 6, but 5 fair dices;

5 fair dice are tossed. What is the probability that at least two of them show the same face.

to asnwer this questions can we use the same formulae:

P(At least 2) = 1 - P(all different)

i think yes, but little unsure..

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Director
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09 Oct 2007, 14:10
Irina, do we know the OA of this question?

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Director
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09 Oct 2007, 14:13
beckee529 wrote:
Irina, do we know the OA of this question?

I copied this question from young_gun at:

http://www.gmatclub.com/forum/p380560

he/she has not provided OA yet...

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Senior Manager
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09 Oct 2007, 20:50
five dice throws.

1-5/6*4/6*3/6*2/6*1/6

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Director
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09 Oct 2007, 21:07
defenestrate wrote:
five dice throws.

1-5/6*4/6*3/6*2/6*1/6

defenestrate,

why do not we start with 6/6 and end with 2/6, instead we start with 5/6 and end with 1/6 ?

for example:

to find number of ways 6 objects can be organized in group of 6 (order matters) 6*5*4*3*2*1

to find number of ways 6 objects can be organized in group of 4 (order matters) 6*5*4*3

from here i thought we should start with 6/6 and end with 2/6...?

Please explain y not, thank you.

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Manager
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11 Oct 2007, 01:32
IrinaOK wrote:
defenestrate wrote:
five dice throws.

1-5/6*4/6*3/6*2/6*1/6

defenestrate,

why do not we start with 6/6 and end with 2/6, instead we start with 5/6 and end with 1/6 ?

Because he is wrong.

6 Dice:
1 - (6!) / 6^6 = 5!/6^5 = 120/6^5 = 1 - 120 / 7776 = 319/324

5 Dice:
1 - (6! / 1!) / 6^5 = 1 - 5!/6^4 = 1 - 120/1296 = 8/9

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Director
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11 Oct 2007, 02:07
JingChan wrote:
IrinaOK wrote:
defenestrate wrote:
five dice throws.

1-5/6*4/6*3/6*2/6*1/6

defenestrate,

why do not we start with 6/6 and end with 2/6, instead we start with 5/6 and end with 1/6 ?

6 Dice:
1 - (6!) / 6^6 = 5!/6^5 = 120/6^5 = 1 - 120 / 7776 = 319/324

5 Dice:
1 - (6! / 1!) / 6^5 = 1 - 5!/6^4 = 1 - 120/1296 = 8/9

Thank you

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Manager
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11 Oct 2007, 03:03
IrinaOK wrote:
defenestrate wrote:
five dice throws.

1-5/6*4/6*3/6*2/6*1/6

defenestrate,

why do not we start with 6/6 and end with 2/6, instead we start with 5/6 and end with 1/6 ?

for example:

to find number of ways 6 objects can be organized in group of 6 (order matters) 6*5*4*3*2*1

to find number of ways 6 objects can be organized in group of 4 (order matters) 6*5*4*3

from here i thought we should start with 6/6 and end with 2/6...?

Please explain y not, thank you.

I think what defenestrate is trying to get at is the fact that 6/6 is just 1, i.e. doesn't matter what number we get in the first roll. What we are concerned with is the fact that this number not show up again, hence starting with 5/6 (the prob of 2nd roll) and so on and so forth. In the case of 6 rolls, the last roll has a prob of 1/6 [5/6*4/6*3/6*2/6*1/6] = 5!/6^5 which is the same as 6!/6^6 if you put 6/6 as the first roll instead of 1.

In the case of 5 rolls, the last roll has a prob of 2/6 [5/6*4/6*3/6*2/6]= 5!/6^4 which is the same as 6!/6^5 if you put 6/6 as the first roll instead of 1.

Additionally, for the sake of completeness, the prob of 5 dice is [1- (120/1296)] = 49/54 not 8/9.

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Director
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11 Oct 2007, 03:21
Skewed,

Thank you for making it clear, really appreciate.

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Manager
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11 Oct 2007, 09:11
Skewed wrote:
Additionally, for the sake of completeness, the prob of 5 dice is [1- (120/1296)] = 49/54 not 8/9.

Whoops, must have made a careless error. =)

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Intern
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Re: PERMUTATION & COMBINATION gurus!!! [#permalink]

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12 Oct 2007, 02:53
IrinaOK wrote:
6 fair dice are tossed. What is the probability that at least two of them show the same face.

Pleas provide your explanations, thank you.

original post available at:

http://www.gmatclub.com/forum/t53592

I found this post about 2 years old and it makes reference to the OA answer:
http://www.gmatclub.com/forum/t21471

OA says (0.985) which makes (1 - 6!/6^6) correct.

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Re: PERMUTATION & COMBINATION gurus!!!   [#permalink] 12 Oct 2007, 02:53
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# 6 fair dice are tossed. What is the probability that at

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