Question

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.

Answer 1

@satellite73

Answer 2

I am having trouble setting it up

Answer 3

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

Answer 4

yes

Answer 5

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

Answer 6

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 ?

Answer 7

no, this is a practice exam

Answer 8

that's ok, thanks for your help

Answer 9

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 \(p_1T_1+p_2T_2\)

Answer 10

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

Answer 11

I am actually having troubling understanding what is being asked

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

probability and schocastic process by yates

Answer 16

nope, sorry

Answer 17

@phi

Answer 18

do you know how to combine two geometric distributions?

Answer 19

no

Answer 20

that is what they are asking

Answer 21

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

Answer 22

can you show me how to combine distributions

Answer 23

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

Answer 24

alright, thanks anyway

Answer 25

if you get the answer, please post it.

Answer 26

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

Answer 27

@zarkon

Answer 28

is that the full problem?

Answer 29

yes

Answer 30

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

Answer 31

screen shot in case I mistyped something

Answer 32

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


\[P(X\le x)=P((X\le x,d1)\text{ or }(X\le x,d2))\]

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

so this is conditional probability problem?

Answer 38

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