Question

Two different numbers are selected simultaneously and at random from the set {1,2,3,4,5,6,7,8,9,10}. What is the probability that the positive difference between the two numbers is 3 or greater

Answer 1

The probability of getting positive difference between the numbers 3 or greater between 1,2,3,4,5,6,7,8,9,10 if two numbers are selected is 28/45.

Given Numbers:{ 1,2,3,4,5,6,7,8,9,10}

We have to find the probability that the positive difference between the numbers selected is 3 or greater.

Probability is the likelihood of happening an event among all the events possible.

Total number of combinations when two numbers are selected from 10 numbers is 10 C 2=10!/2!(10-2)!

=10*9*8!/2*1(8!)

=10*9/2

=45 combinations

Combinations of numbers whose difference will be 3 or more are: (10,7),(10,6),(10,5),(10,4)(10,3)(10,2)(10,1)(9,6)(9,5)(9,4)(9,3)(9,2)(9,1)(8,5)(8,4)(8,3)(8,2)(8,1)(7,4)(7,3)(7,2)(7,1)(6,3)(6,2)(6,1)(5,2)(5,1)(4,1)

Total numbers =28

Probability=numbers/total combinations

=28/45

Hence the probability that two numbers selected gives difference of 3 or more is 28/45.

Learn more about probability at brainly.com/question/24756209

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