A production line yields two types of devices:
Type1 devices occur with probability p1 and work for an average time T1,
Type2 devices occur with probability p2 and work for an average time T2,
such that p1+p2=1 and T2>T1.
The time a device works is modeled as a geometrical distribution.
Let X be the time a device chosen uniformly at random works.
Find the distribution of X.

- anonymous

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- anonymous

@satellite73

- anonymous

I am having trouble setting it up

- anonymous

ok i have to think
geometric distribution is \(P(n) = p(1-p)^n\) right?

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## More answers

- anonymous

yes

- anonymous

jesus i have no idea. it tell you average is \(T_1\) and i think that means \(\frac{1-p_1}{p_1}=T_1\)

- anonymous

let me try to write this correctly
\[T_1=\frac{p_2}{p_1}\] and \[T_2=\frac{p_1}{p_2}\] but i am not sure that helps solve the problem
do you have an example to work off of ?

- anonymous

no, this is a practice exam

- anonymous

that's ok, thanks for your help

- anonymous

ok forget that last noise i wrote, it is ridiculous and i am embarrassed
maybe it is much simpler
we pick device \(T_1\) with probability \(p_1\) and device \(T_2\) with probability \(p_2\)
could it be as simple as \(p_1T_1+p_2T_2\)

- anonymous

no that isn't right either, at least i don't think so
what text are you using?

- anonymous

I am actually having troubling understanding what is being asked

- anonymous

i think you are being asked the following:
pick a device at random, put \(X\) = amount of time it runs.
Find the distribution of \(X\)

- anonymous

in other words, find a formula for \(P(X=k)\) the probability the device runs for time \(k\)

- anonymous

but really i am stuck so i should shut up. but perhaps i have a text with something similar, which is why i asked what text you were using

- anonymous

probability and schocastic process by yates

- anonymous

nope, sorry

- anonymous

@phi

- phi

do you know how to combine two geometric distributions?

- anonymous

no

- phi

that is what they are asking

- anonymous

i should delete my embarrassing answer
now we will get a real one i hope

- anonymous

can you show me how to combine distributions

- phi

No answer from me, I never learned this stuff. But it might be a negative binomial distribution.

- anonymous

alright, thanks anyway

- phi

if you get the answer, please post it.

- anonymous

I will ask zarkon, when he comes on; hopefully before my quiz though

- anonymous

@zarkon

- Zarkon

is that the full problem?

- anonymous

yes

- Zarkon

I'm a little confused ...
Type1 devices occur with probability p1 ...
Type2 devices occur with probability p2
vs
"a device chosen uniformly"
are they chosen uniformly or with probabilities p1,p2

- anonymous

screen shot in case I mistyped something

##### 1 Attachment

- Zarkon

I don't think the problem is worded very well especally near the end.
here is how I see the problem
let \(d1,d2\) be device 1 and device 2
then
\[P(X\le x)=P((X\le x,d1)\text{ or }(X\le x,d2))\]
\[=P(X\le x|d1)P(d1)+P(X\le x|d2)P(d2)\]

- Zarkon

\[=P(X\le x|d1)p1+P(X\le x|d2)p2\]

- Zarkon

though I could be misinterpreting the problem (since I don't think it is totally clear)

- anonymous

yeah, I had hard time even understanding what they were asking for

- Zarkon

I would ask your prof...problem isn't clear to me and I have taught courses in mathematical statistics.

- anonymous

so this is conditional probability problem?

- Zarkon

based on what read it looks like you need to use conditional prob for at least part of the problem

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