# Lightbulb Probability Density Function Problem

#### oglemetender

##### New member
I have been trying to solve a problem about the probability that 3 lightbulbs will last for 1500 hours of operation if the probability density function of a lightbulb is f(x) = 6[.25-(x-1.5)^2] when 1 <= x <= 2 and f(x) = 0 otherwise. x is measured in multiples of 1,000 hours. I tried doing [the integral from 1 to 1.5 of f(x)] CUBED, but I got an answer greater than 1. Does anyone know what I am doing wrong?

#### tkhunny

##### Moderator
Staff member
Always keep track of yourself.

$$\displaystyle \int_{1}^{2}6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\;dx\;=\;1$$

This is a proper pdf

When you apply the exponent, 3, you get...

$$\displaystyle \int_{1}^{2}\left{6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\right}^{3}\;dx\;=\;\frac{54}{35}\;>\;1$$

Clearly, this is no good.

Try this:

$$\displaystyle \int_{1.5}^{2}6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\;dx\;=\;\frac{1}{2}$$

After that, it's a Binomial problem with p = 1/2 and n = 3. Let's see what you get.