Find the probablity of getting a sum of at least 8 by tossing a six-sided die twice.
I keep getting 5/36 but my book says 5/12.
Of the 36 possibilities, when tossing the same die twice, the ones that give a sum of 8 or more are:
6,2; 6,3; 6,4; 6,5; 6,6
5,3, 5,4, 5,5, 5,6
4,4; 4,5; 4,6
3,5; 3,6
2,6
That is 15 out of 36, for a 5/12 probability of 8 or more.
That is also the probability of tossing eight or more with two dice.
To find the probability of getting a sum of at least 8 by tossing a six-sided die twice, we need to count the number of favorable outcomes and divide it by the total number of possible outcomes.
First, let's find the total number of possible outcomes. When you roll a six-sided die twice, there are 6 possible outcomes for each roll. Since we are rolling the die twice, we multiply the number of outcomes for each roll together. Therefore, there are 6 * 6 = 36 possible outcomes.
Next, let's determine the number of favorable outcomes. To get a sum of at least 8, we need to consider the possible combinations of the two rolls that will result in a sum of 8, 9, 10, 11, or 12.
To calculate this, we can create a table:
+-------+------+------+------+------+------+
| Roll | 1st | 2nd | 3rd | 4th | 5th |
+-------+------+------+------+------+------+
| First | 2 | 3 | 4 | 5 | 6 |
| Second| 6 | 5 | 4 | 3 | 2 |
+-------+------+------+------+------+------+
From the table, we can see that there are 5 favorable outcomes. Therefore, the probability of getting a sum of at least 8 is 5/36.
Now, the issue you mentioned is that your book says the probability is 5/12. This is incorrect. Based on the analysis we just did, the correct probability is indeed 5/36.
It's possible that your book made an error or there might be some differences in the problem statement given in the book. It's always important to double-check the problem statement and calculations to ensure accuracy.