# Probability f(x) = (2/9)(3x -x^2)

## Homework Statement

Find the probability P(x>1)
The function is f(x) = (2/9)(3x -x^2)

## The Attempt at a Solution

So,
P(x>1) = ∫ (2/9)(3x-x^2) dx = (2/9)((3x^2/2) - (x^3/3))[1,∞]

When you do the improper integral lim a->∞ [ (2/9)((3(a)^2/2) -((a)^3)/3)] this is just the first part of the evaluation I stopped because it goes to infinity. So If the integral does not converge then I say the event will never happen?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find the probability P(x>1)
The function is f(x) = (2/9)(3x -x^2)

## The Attempt at a Solution

So,
P(x>1) = ∫ (2/9)(3x-x^2) dx = (2/9)((3x^2/2) - (x^3/3))[1,∞]

When you do the improper integral lim a->∞ [ (2/9)((3(a)^2/2) -((a)^3)/3)] this is just the first part of the evaluation I stopped because it goes to infinity. So If the integral does not converge then I say the event will never happen?

Surely this cannot be the whole problem statement, because it is nonsense as written. Do you mean that f(x) is the probability density function? If so, say it. But that is a rather minor quibble. More serious the fact that for some value of x, f(x) becomes < 0 and that can never, ever happen for a legitimate probability density. Also: the density is supposed to integrate to 1, but yours gives a divergent integral.

So, you need to tell us EXACTLY what the problem says.

Problem says.
f(x) = k(3x-x^2) if 0<= x <= 3 and f(x) = 0 if x < 0 or x>3
a. for what value of k is f a probability density funciton?
b. for that value of k find P(X>1)
c. Find the mean.
I'm asking about b. K was determined to be 2/9

Yeah I found the mistake it is [0,3] not infinity. So I can do integrate from [0,1] then do 1- whatever the result is.
to get P