Question

The geometric probability function is f (x) = (1-P) x-1 P. what is the approximate probability of rolling a standard die and getting the first 6 on the 3rd try?​

Answer 1

We know that for a standard dice the probability of obtain a 6 is:

`P=(1)/(6)`

And for this case our value of x=3 and replacing we got:

`f(x=3) = (1- (1)/(6))^(3-1) (1)/(6)`

`f(x=3)=(25)/(36) (1)/(6)= (25)/(216)= 0.116`

Explanation:

For this case we have the following function:

`f(x) = (1-P)^(x-1) P`

We want to find the approximate probability of rolling a standard die and getting the first 6 on the 3rd try

We know that for a standard dice the probability of obtain a 6 is:

`P=(1)/(6)`

And for this case our value of x=3 and replacing we got:

`f(x=3) = (1- (1)/(6))^(3-1) (1)/(6)`

`f(x=3)=(25)/(36) (1)/(6)= (25)/(216)= 0.116`