Question

Two numbers are randomly selected on a number line numbered 1 through 9. The same number can be chosen twice. What is the probability that both numbers are greater than 6?

19

12

13

16

Answer 1

I'm not the best when it comes to probability but I think I can solve this.
First of all, we have to calculate al lthe possible outcomes, these being all the possible permutations of selections, these include the double-scenarios.
So, the possible outcomes are the ordered pairs we can form : \[(1,1)-(1,2)-(1,3)-...-(9,1)-(9,2)-...(9,9)\]
This means there are \(9^2\) possible outcomes of this scenario BUT under the given conditions we will have to establish all the scenarios where the numbers will be greater than 6, this means from 6 to 9 there are three numbers and all the permutations possible are \(3^2\). Let's call it the probability "P" and it will be composed by the division of the allowed permutations with the entirety of all the possible outcomes:
\[P=\frac{ 3^2 }{ 9^2 }\]
And you should be able to take over from here.

Answer 2

thank you!