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.

MathCuber
Jul 4, 2018

#1**+1 **

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.**

Guest Jul 4, 2018

#2**+1 **

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 \(\frac{30}{49}\)

Melody
Jul 5, 2018