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

Question

(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?

javropu · Answer

mmmmm, lets see...
1. total number of possibilities=6*6*6/3!
number satisfying 2 equal numbers=6*1*5/3!
answer: 5/36
2. total number of possibilities=6^4/4!
number satisfying 0 equal numbers=6*5*4*3/4!
answer: 10/36
3. total number of possibilities=6^6/6!
there is only one way of having different numbers, dont need to calculate that
answer: 1/(6^6/6!)

stolyar · Answer

my answers:

(1) 5/36
(2) 5/18
(3) 5/324

kheriapiyush@indiatimes.c · Answer

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.

mciatto · Answer

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.

Comments?

MBA04 · Answer

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

mciatto · Answer

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

Brainless · Answer

(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

stolyar · Answer

My approaches are the very same.

stolyar · Answer

(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?

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

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