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anonymous
 3 years ago
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
 3 years ago
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am having trouble setting it up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i have to think geometric distribution is \(P(n) = p(1p)^n\) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0jesus i have no idea. it tell you average is \(T_1\) and i think that means \(\frac{1p_1}{p_1}=T_1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let 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
 3 years ago
Best ResponseYou've already chosen the best response.0no, this is a practice exam

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's ok, thanks for your help

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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
 3 years ago
Best ResponseYou've already chosen the best response.0no that isn't right either, at least i don't think so what text are you using?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am actually having troubling understanding what is being asked

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0in other words, find a formula for \(P(X=k)\) the probability the device runs for time \(k\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but 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
 3 years ago
Best ResponseYou've already chosen the best response.0probability and schocastic process by yates

phi
 3 years ago
Best ResponseYou've already chosen the best response.0do you know how to combine two geometric distributions?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0that is what they are asking

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i should delete my embarrassing answer now we will get a real one i hope

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you show me how to combine distributions

phi
 3 years ago
Best ResponseYou've already chosen the best response.0No answer from me, I never learned this stuff. But it might be a negative binomial distribution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright, thanks anyway

phi
 3 years ago
Best ResponseYou've already chosen the best response.0if you get the answer, please post it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I will ask zarkon, when he comes on; hopefully before my quiz though

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0is that the full problem?

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0I'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
 3 years ago
Best ResponseYou've already chosen the best response.0screen shot in case I mistyped something

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 \[P(X\le x)=P((X\le x,d1)\text{ or }(X\le x,d2))\] \[=P(X\le xd1)P(d1)+P(X\le xd2)P(d2)\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0\[=P(X\le xd1)p1+P(X\le xd2)p2\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0though I could be misinterpreting the problem (since I don't think it is totally clear)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, I had hard time even understanding what they were asking for

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0I would ask your prof...problem isn't clear to me and I have taught courses in mathematical statistics.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so this is conditional probability problem?

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0based on what read it looks like you need to use conditional prob for at least part of the problem
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