Probability Questions with Multiple Dices

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I came across this problem in a book that I bought recently. Three dices are dropped. Find the probability that when we roll the dices:

a) In all the dices we score 5. (Not as a sum) My answer:3/18 = 1/6

b) In at least one dice we score 6 My answer : 3/18=1/6

c) The sum of the scores in the dices is smaller or equal to six My answer :
we have the combinations (1 1 1), (1 1 2), (1 1 3),(1 1 4),(2 1 2), (2 1 3), (4 1 1) so my answer is 7/18

d) The sum of the scores in the dices is bigger than the product of the scores..I don't know how to do this one

e) In none of the dices we score 5 or 6.This has to be 3/18+3/18 =6/18=1/3

f) In two dices we have the same score..I don't know how to do this one either :/
 
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Let's start with the first one. Can you explain your answer for a) ?

How many possible outcomes are there if you throw three dice?
 
By the way- "dice" is itself the plural of the word "die". There is no such word as "dices".
 
Elaia06 said:
I came across this problem in a book that I bought recently. Three dices are dropped. Find the probability that when we roll the dices:

a) In all the dices we score 5. (Not as a sum) My answer:3/18 = 1/6

b) In at least one dice we score 6 My answer : 3/18=1/6

c) The sum of the scores in the dices is smaller or equal to six My answer :
we have the combinations (1 1 1), (1 1 2), (1 1 3),(1 1 4),(2 1 2), (2 1 3), (4 1 1) so my answer is 7/18

d) The sum of the scores in the dices is bigger than the product of the scores..I don't know how to do this one

e) In none of the dices we score 5 or 6.This has to be 3/18+3/18 =6/18=1/3

f) In two dices we have the same score..I don't know how to do this one either :/

It is *very* important that you show us your work, at least for (a) and (b), because I cannot figure out how you could possibly get the types of numbers you have written. Where does the denominator '18' come from?
 
Ray, I would guess it is 3 * 6, which is an understandable error but incorrect.
 
CompuChip said:
Ray, I would guess it is 3 * 6, which is an understandable error but incorrect.

I knew that, of course, but I wanted the OP to spell it out. No luck so far.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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