Question

In a "torture test" a light switch is turned on and off until it fails. If the probability that the switch will fail anytime it is turned on or off is .001
, what is the probability that the switch will fail after it has been turned on or off 1,200
times? Assume that the conditions underlying the geometric distributions are met. [Hint:
use the formula for the value of an infinite geometric progression]

Answer 1

I'm a bit unsure of how to use the infinite geometric progression formula in order to solve this. From what I've seen online:

http://www.proofwiki.org/wiki/Sum_of_Infinite_Geometric_Progression#Theorem
\[\Sigma|z|^{n}\frac{ 1 }{ 1-|z| }\]
The sigma should read from n=0 to infinity.

Answer 2

I would use
http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series

\[\sum_{k=0}^{n}ar^k=\frac{a (1-r^{n+1})}{1-r} \]

Answer 3

the prob it fails on the first test is 0.001
the prob it fails on the 2nd test is
0.999* 0.001 (prob it passes the first test times prob of failing on the second)
you get a series
0.001 + 0.999*0.001 + 0.999^2 * 0.001 + 0.999^3 * 0.001 + ...

Answer 4

Alright. I think I've just figured it out. Thanks.

Answer 5

I think the answer is 0.001 * (1-0.999^1200)/(1-0.999)
I used n= 1199 (0 to 1199 means 1200 tests)

Answer 6

yep thats what I did and I ended up getting, .3010