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# 1) One has three fair dice and rolls them together. What is

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1) One has three fair dice and rolls them together. What is [#permalink]

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17 Aug 2003, 01:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

(1) One has three fair dice and rolls them together. What is the probability of having two same numbers and one different? (for example, 2-2-6, 1-3-3, 6-5-6, and so on)

(2) One has four fair dice and rolls them together. What is the probability of having all the numbers different? (for example, 1-2-3-4, 6-5-4-1, and so on)

(3) One has six fair dice and rolls them together. What is the probability of having all the numbers different?
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17 Aug 2003, 01:52
mmmmm, lets see...
1. total number of possibilities=6*6*6/3!
number satisfying 2 equal numbers=6*1*5/3!
2. total number of possibilities=6^4/4!
number satisfying 0 equal numbers=6*5*4*3/4!
3. total number of possibilities=6^6/6!
there is only one way of having different numbers, dont need to calculate that
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17 Aug 2003, 07:47

(1) 5/36
(2) 5/18
(3) 5/324
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18 Aug 2003, 02:06
I would like to have the explanation for the 1st one.

I tried the following:-
1) Total no. of possibilities = 6*6*6 = 216
No. of events where all the three are equal = 6
No. of events when none are eqaul = 6*5*4 = 120
Therefore the no. of events when two are equal and third different
= 216-6-120 = 90

Therefore Probability that 2 are equal and third is different = 90 / 216
= 5/12

The remaining two I got the answers same as 5/18 and 5/324 respectively.
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18 Aug 2003, 06:25
kheriapiyush@indiatimes.c wrote:
I would like to have the explanation for the 1st one.

I tried the following:-
1) Total no. of possibilities = 6*6*6 = 216
No. of events where all the three are equal = 6
No. of events when none are eqaul = 6*5*4 = 120
Therefore the no. of events when two are equal and third different
= 216-6-120 = 90

Therefore Probability that 2 are equal and third is different = 90 / 216
= 5/12

The remaining two I got the answers same as 5/18 and 5/324 respectively.

I agree with indiatimes on this one. My method was the same:

All three numbers the same we have 6/216.
All three different we have 120/216.
So the remainder is 90/216, reducing to 5/12.

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18 Aug 2003, 06:39
My aproach for the 1st is

first dice can be any number. Second dice must be the same (order does not matter) 1/6. Third dice must be any number but not the same as in 1 and 2, so 5/6

1/6*5/6= 5/36

With number 2 same aproach

5/6*4/6*3/6 = 5/18

With number three same aproach

5/6*4/6*3/6*2/6*1/6 = 5/324
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18 Aug 2003, 07:00
Maybe its just b/c its Monday morning (I hope!!)...

Can sombody please break down there three possibilities of part (A)..

Probability that...

(1) All three are the same
(2) All three are different
(3) Two are the same, and 1 is different
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20 Aug 2003, 13:25
(1) As explained by Stolyar we have three mutually exclusive situations

(A) S S D ,
(B) S D S ,
(C) D S S

where S = Same number , D = Different Number

p(A) = 1/6 x 1 x 5/6 = 5/36
p(B) = 1/6 x 5/6 x 1 = 5/36
p(C) = 5/6 x 1/6 x 1 = 5/36

Combined Probability = p(A) + p(B) + p(C) = 3 x 5 / 36 = 5/12
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30 Aug 2003, 23:34
MBA04 wrote:
My aproach for the 1st is

first dice can be any number. Second dice must be the same (order does not matter) 1/6. Third dice must be any number but not the same as in 1 and 2, so 5/6

1/6*5/6= 5/36

With number 2 same aproach

5/6*4/6*3/6 = 5/18

With number three same aproach

5/6*4/6*3/6*2/6*1/6 = 5/324

My approaches are the very same.
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30 Aug 2003, 23:41
mciatto wrote:
Maybe its just b/c its Monday morning (I hope!!)...

Can sombody please break down there three possibilities of part (A)..

Probability that...

(1) All three are the same
(2) All three are different
(3) Two are the same, and 1 is different

(1) 1*1/6*1/6=1/36 (all the three are the same)
(2) 1*5/6*4/6=20/36=5/9 (all the three are different)
(3) 1*1/6*5/6=5/36 (2+1)

Another interesting question:
1/36+20/36+5/36=26/36
the full set of probabilities is 1, so 10/36 is the probability of what?
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