Exponential probability density

[EV33]

Homework Statement

a.) A lamp has two bulbs of a type with an average lifetime of 1000 hours. Assuming that we can model the probability of failure of these bulbs by an exponential density function with mean u = 1000, find the probability that both of the lamp's bulbs fail within 1000 hours.
b.) Another lamp has just one bulb of the same type as in part (a). If one bulb burns out and is replaced by a bulb of the same type, find the probability that the two bulbs fail within a total of 1000 hours.

Homework Equations

Exponential density function: f(t) = { 0 if t < 0
u^-1e^-t/u if t ≥ 0}

The Attempt at a Solution

Plugging 1000 in for u, I got f(t) = { 0 if t < 0
1000^-1e^-t/1000 if t ≥ 0}

My guess is that I need to say that the function I got is the function for both light bulb #1 (I'll call it X) and light bulb #2 (I'll call it Y). Therefore, I am assuming I need to multiply these two functions X and Y together and then find when P(X+Y ≤ 1000). Therefore I would integrate the function where x goes from 0 to 1000 and y goes from 0 to 1000 - x.

Am I right in my steps cause I am really not sure if I completely understand probability density...?

Thanks.

 

[EV33]

Can anyone tell me if my steps are right?

 

[LCKurtz]

You could have each bulb fail at 900 hours so X+Y = 1800. So X + Y < 1000 isn't what you are wanting.

 

[EV33]

So would it be X + Y < 2000 then? So my bounds would be from 0 to 2000 - x and 0 to 2000?

 

[LCKurtz]

No. Why are you thinking about X + Y in the first place? The question isn't about a sum. You have two events, presumably independent although you didn't say so, that must both happen.

 

[EV33]

I am thinking about X + Y cause I have no idea how to approach this problem so I was just copying the other example of an exponential density function in my book.

 

[vela]

Which part are you trying to solve, a or b?

 

[LCKurtz]

Which part are you trying to solve, a or b?

Good question; I'm assuming (a).