Two different numbers are chosen from the set {1, 2, 3, ..., 10}. What is the probability that their sum is 8 or greater?
+1084 First, let's decide what the denomintor will be. In this case, it is 10 times 9, which is 90. Since there can be repeats of the same group, we devide by 2!, which is 2, and that will give us 90
From here, let's see what combos will end up being greater than or equal to 8 by using a table:
| Number. | 1. | 2 | 3. | 4. | 5. | 6. | 7. | 8. | 9. | 10 | Total: |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 4 |
| 2 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 5 |
| 3 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 6 |
| 4 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 6 |
| 5 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 7 |
| 6 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 8 |
| 7 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 9 |
| 8 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 9 |
| 9 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 9 |
| 10 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 9 |
Now, we have to add all the totals up. This is 4+5+6(2)+7+8+4(9) = 72. Dividing by 2, we get 36. Therefore, the probability is 36/45, which is 4/5.
The following are number of ways you can choose 2 numbers out of 10:
{1, 2} | {1, 3} | {1, 4} | {1, 5} | {1, 6} | {1, 7} | {1, 8} | {1, 9} | {1, 10} | {2, 3} | {2, 4} | {2, 5} | {2, 6} | {2, 7} | {2, 8} | {2, 9} | {2, 10} | {3, 4} | {3, 5} | {3, 6} | {3, 7} | {3, 8} | {3, 9} | {3, 10} | {4, 5} | {4, 6} | {4, 7} | {4, 8} | {4, 9} | {4, 10} | {5, 6} | {5, 7} | {5, 8} | {5, 9} | {5, 10} | {6, 7} | {6, 8} | {6, 9} | {6, 10} | {7, 8} | {7, 9} | {7, 10} | {8, 9} | {8, 10} | {9, 10} (total: 45)
Their sums =(3, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 17, 17, 18, 19)>Total = 45
There are 39 out of 45 which are 8 and >. Therefore, the probability is: 39 / 45
