Question 1197527: An experiment consists of rolling two fair (not weighted) dice and adding the dots on the two sides facing up. Each die has the number 1 on two opposite faces, the number 2 on two opposite faces, and the number 3 on two opposite faces. Compute the probability of obtaining the indicated sum.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52788)   (Show Source):
You can put this solution on YOUR website! .
An experiment consists of rolling two fair (not weighted) dice and adding the dots on the two sides facing up.
Each die has the number 1 on two opposite faces, the number 2 on two opposite faces, and the number 3
on two opposite faces. Compute the probability of obtaining the indicated sum.
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I bring it to your attention that there is no indicated sum in this post.
Or, may be, it is a secret info and is written by invisible font ?
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The only numbers on the dice are 1, 2, and 3; there are 2 of each number on each die, so the probability of each number on each die is 1/3. Note the information that the two 1's are on opposite faces, as are the two 2's and the two 3's, is not relevant to the question.
Since the probability of each number on each die is 1/3, the probability of any particular ordered outcome is (1/3)(1/3)=1/9.
Your post did not indicate a particular sum; but since there are not many possible sums it is easy to calculate the probability of each.
sum of 2:
1+1; number of outcomes with this sum: 1
probability(sum of 2) = 1(1/9) = 1/9
sum of 3:
1+2 or 2+1; number of outcomes with this sum: 2
probability(sum of 3) = 2(1/9) = 2/9
sum of 4:
1+3, 2+2, or 3+1; number of outcomes with this sum: 3
probability(sum of 4) = 3(1/9) = 3/9 = 1/3
sum of 5:
2+3 or 3+2; number of outcomes with this sum: 2
probability(sum of 5) = 2(1/9) = 2/9
sum of 6:
3+3; number of outcomes with this sum: 1
probability(sum of 6) = 1(1/9) = 1/9
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