+50 Suppose we take two fair dice, but change the numbers on the faces. On one die, we'll replace the 2 with a 5, and on the other, we'll replace the 5 with a 2. What is the probability that the sum of the numbers are shown when both are rolled is 7? I tried some things but Im just confused on the part where it says replacing a number on the dice with another one. i need a full solution, not answers because i want to actually understand the problem not just put in the answers and be done with it. dispite my username.
A standard die has 6 faces on it, 1, 2, 3, 4, 5, 6, right? In the case of your 2 dice, they have changed like this:
First die ==1, 3, 4, 5, 5, 6 [It has no 2 on it, but 2 fives]
2nd die ==1, 2, 2, 3, 4, 6 [It has no 5 on it, but 2 twos]. OK, so far?
Form an additions table as follows:
1, 3, 4, 5, 5, 6
1
2
2
3
4
6
If you make an additions table of 6 X 6 numbers (as is the case with your 2 dice), then you should have 6 * 6 ==36 sums as follows:
(2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 12)==36 sums
Count the number of "7s" from the above sums and you get ==8 "sevens"
Therefore, the probability of getting a sum of 7, when rolling your 2 "altered" dice is: 8 /36 ==2 / 9
+876 You have: $(1, 3, 4, 5, 5, 6), (1, 2, 2, 3, 4, 5, 6)$
For $1,$ you can only have $6,$ and there is $1$ of them, so $1$ way to do this.
For $3,$ you can only have $4,$ and there is $1$ of them, so $1$ way to do this.
For $4,$ you can only have $3,$ and there is $1$ of them, so $1$ way to do this.
For both of the $5,$ you can only have $2,$ and there are $2$ of them, so $2 \times 2 = 4$ way to do this.
For $6,$ you can only have $1,$ and there is $1$ of them, so $1$ way to do this.
In total, there are $1 + 1 + 1 + 4 + 1 = \boxed{8}$
