Question

Need help with a probability calc question

Answer 1

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 \[\mu = 1000\], find the probability that both of the lamp's bulbs fail within 1000 hours.

Answer 2

Really just need to know how to set up a probability density function for a bulb

Answer 3

Sure, so what kinds of stuff do you already know about exponential density functions? Like, tell me what mu is for example if that's all you know, or do you think maybe the graph is going to be increasing or decreasing or what sort of behavior are you thinking about here?

Answer 4

I don't really know of exponential density function, as I've never used it before, I guess that's all I need and I can do the rest haha. It's kind of hard to tell how the region will be like at the moment.

Answer 5

Mui is the mean of the probability function

Answer 6

i have a suspicion it involves a double integral

Answer 7

Yup!

Answer 8

are the bulbs life independent?

Answer 9

Yes

Answer 10

This might work
P( x < 1000) = 1-e^-(1000/1000) =0.63212
For 2 bulbs, assuming independence = (0.63212)(0.63212)

Answer 11

Hey, could you explain how you got that?

Answer 12

I don't think we could use Poisson here.

Answer 13

\[f_{X}(x)=\frac{1}{\mu}e^{-x/\mu}\]
\[\int\limits_{0}^{t}f_{X}(x)dx=1-e^{-t/\mu}\]

Answer 14

Oh wait, I think I got it, I just needed to know the exponential density function as I mentioned before haha, which is \[\huge f(x) = \frac{ 1 }{ \mu } e^{\frac{ -x }{ \mu }}\]

Answer 15

Yeah, I see now :)

Answer 16

\[f(x) = \frac{ e^{-x/1000} }{ 1000 }~~~x \ge 0\] same goes for f(y) so it'll be \[P(0 \le x \le 1000, 0 \le y \le 1000) \] and the double integral is pretty easy from there

Answer 17

Want me to show you how to derive the probability density function and the average? It's completely intuitive.

Answer 18

Yes! That would be very useful :)

Answer 19

Ok, so I'm going to be super relaxed on some stuff so don't expect too much rigor out of this, so here we go! First off I'm going to use the greek letter rho to denote density, but now it's probability density! That means it's amount of probability per distance, time, or whatever parameter your probability varies over! \[\Large \rho(x) = \frac{P(x)}{dx}\] Notice that this is not a derivative the way I've written it, so don't beat me up or anything it's not rigorous I said lol.

Answer 20

We can find the probability of the event happening by integrating our probability density over a span of time, like from 1 minute after installing it to tomorrow and that value we get out of the integral will tell us the probability that the lightbulb will go out in that time. Hopefully it's a small number.So we know that our probability has to evaluate to 1 if we look at all the possible values our probability can take, so we set up this integral: \[\Large \int\limits_{-\infty}^ {\infty} \rho(t) dt = 1 \] Ok so this specific probability density is over time, so we will set the first bound on the integral to 0 instead of negative infinity since we are using a new lightbulb at t=0. I don't think we are considering lightbulbs that were created at the beginning of the universe or stuff like that, so this makes sense. Another thing, we said exponential didn't we? So let's plug that in now. \[\Large \rho(t) = A e^{-bt}\] Two random constants, we don't know what they are yet. That's fine, we'll figure it out soon. \[\Large \int\limits_0^\infty A e^{-bt}dt = 1\]

Answer 21

Integrate that, we get A=b and now we have to talk about expectation values lol... Ok this ended up being a lot longer than I expected in my mind it was shorter.

Answer 22

Lol it's cool, pretty easy to follow.

Answer 23

Alright so now if we have a bunch of objects how do we find the average? \[\Large \frac{1+1+1+1+4+4}{6}=2=avg\] Well alternatively we can rearrange the formula like this: \[\Large \frac{4*1+ 0*2+0*3+2*4}{6}\] I inserted a 0 to show that I am giving a weight to each value, there are 4 ones, 0 twos, 0 threes, and 2 fours. This is really just our density function! If we evaluate a density function we get the weight. rho(2)=0 and rho(4)=2 for example. Now we can rewrite this as: \[\Large \frac{1}{6} \sum_{n=1}^4 n*\rho(n)\] So the 6 on bottom is the total number of numbers. In discrete terms we already know (and I probably should have started with the discrete to begin with, but oh well) \[\Large \frac{1}{6} \sum_{n=1}^4 \rho(n)=1\] See how this works out, all the weights added up represent all the numbers, so dividing the total number of number weights by the total number of numbers is going to always be 1, maybe I didn't explain that quite right, but it might make sense.

Answer 24

So that discrete sum I showed first with n*rho(n) is the average, also known as the expected value. So we can see that our integral will look like this: \[\Large \mu = \int\limits_0^\infty t* be^{-bt}dt\] Yeah you have to do integration by parts, but if you do it you will be satisfied to see that \[\Large b= \frac{1}{\mu}\] just like you plugged in earlier, isn't that nice?

Answer 25

Of course, this isn't the density function you really want, since it's nonsense by the way for your model.

Answer 26

You'll want something that starts out small and gets larger not starts out large and gets smaller. The probability of a lightbulb burning out gets more likely as time goes on, doesn't it? So you'll want to have your exponential function slightly tweaked to look like this: