+537 Two different numbers are selected simultaneously and at random from the set {1, 2, 3, 4, 5, 6, 7}. What is the probability that the positive difference between the two numbers is 2 or greater? Express your answer as a common fraction.
I will give this a try:
Since you have a set of {1, 2, 3, 4, 5, 6, 7} permuated 2 at a time, then you have a total of: 7P2 = 42 permutations as follows:
{1, 2} | {1, 3} | {1, 4} | {1, 5} | {1, 6} | {1, 7} | {2, 1} | {2, 3} | {2, 4} | {2, 5} | {2, 6} | {2, 7} | {3, 1} | {3, 2} | {3, 4} | {3, 5} | {3, 6} | {3, 7} | {4, 1} | {4, 2} | {4, 3} | {4, 5} | {4, 6} | {4, 7} | {5, 1} | {5, 2} | {5, 3} | {5, 4} | {5, 6} | {5, 7} | {6, 1} | {6, 2} | {6, 3} | {6, 4} | {6, 5} | {6, 7} | {7, 1} | {7, 2} | {7, 3} | {7, 4} | {7, 5} | {7, 6} (total: 42)
I count 15 permutations of 2 numbers out of a total of 42 that the positive difference between those 15 numbers will be 2 or greater. Therefore, the probability is:
15 / 42 = 5 / 14
Note: Somebody should check this.
+118609 Two different numbers are selected simultaneously and at random from the set {1, 2, 3, 4, 5, 6, 7}. What is the probability that the positive difference between the two numbers is 2 or greater? Express your answer as a common fraction.
If they are selected simultaneously then they could be the same number.
So there are 7*7 = 49 possible pairs.
Ho many DO NOT have a difference or 2 or more.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | |
| 1 | 0 | 1 | |||||
| 2 | 1 | 0 | 1 | ||||
| 3 | 1 | 0 | 1 | ||||
| 4 | 1 | 0 | 1 | ||||
| 5 | 1 | 0 | 1 | ||||
| 6 | 1 | 0 | 1 | ||||
| 7 | 1 | 0 |
That is 19 with a diff of 2 or less.
49-19 = 30
so the prob of the difference being 2 or greater is \(\frac{30}{49}\)
