On Sep 29, 2011, at 4:02 AM, Man on Bridges wrote:

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Hi, On 29-9-2011 8:27, Horace Heffner wrote:Looking at the other side of the coin, the probability ofcatastrophic failure, suppose there is a 0.1% chance per hour oneof the E-cats can blow up spreading steam throughout thecontainer. There is thus a 0.999 probability of success, i.e. noexplosion for one E-cat, operating for one hour. Theprobability that all 52 E-cats perform successfully for a 24 hourtest period is then 0.999^(52*24) = .287. That means there is a71.3% chance of an explosion during a 24 hour test.Me thinks you are wrong. Your statistical probability calculationis based upon the fact that the chance of a single Ecat explodingis influenced by it's behaviour earlier,

`This is false. The probability in each time increment is assumed to`

`be independent. For there to be success there must be no failures for`

`any time increment. If there are T time increments, and the`

`probability of failure in any time increment is p, the probability of`

`success q=1-p in each time increment is independent of the other time`

`increments, and the probability of success in all time increments is`

`q^T (only possible if what happens in each time increment is`

`independent event), and the probability of any failure having`

`occurred is thus 1-(q^T).`

which of course is not true. Statistically each Ecat has it's ownindependent chance of explosion at any given moment which does notchange over time.

`The instantaneous probability of failure is zero. Zero time results`

`in zero probability because lim t->0 q^t = 1 for for all 0=<q<=1 and`

`positive t. Therefore lim t->0 1-(q^t) = 0. Note that I provided an`

`assumption of 0.001 percent probability of failure *per hour*.`

With your probability of 0,1% chance per hour this would result forthe whole of 52 Ecats then in a chance of explosion at any givenmoment of 1 - (0.999^52) = .05 or 5%.

`No. The probability of at least one E-cat failure in the 52 E-cat`

`system, based on the assumption of 0.001 probability of failure of an`

`individual E-cat in an hour is 1-(0.999)^52 = 0.506958 = 5%. Your`

`number 5% is right, but your interpretation of it representing an`

`instantaneous moment is wrong.`

Looking even a bit more closer again this would mean that if thechance of explosion is 0.1% per hour then the chance of explosionis 2,77e-7 per second at any given moment for a single Ecat, whichwould result for 52 Ecats into 1-((2,77e-7)^52) = 0,00001444434 or0,00144% at any time.

The phrase "at any time" makes the above statement nonsensical.

`An hour represents 3600 seconds, which are 3600 independent events of`

`1 second duration. Let a be the probability of failure in 1 second,`

`and b=(1-a) be the probability of success in 1 second. We have the`

`given probability p of failure for 3600 seconds being 0.001, and the`

`probability of success of one E-cat for one hour being q = 0.999.`

`The probability of success (no failures) for the 3600 1 second`

`independent time increments is`

q = 0.999 = b^3600 b = q^(1/3600) = 0.999^(1/3600) a = 1 - 0.999^(1/3600) = 2.779x10^-7

`Note that a is the probability of failure in one second, not "at any`

`time". This is totally consistent with the probability of failure in`

`one E-cat in one hour being 5%. In other words, going backwards:`

p = 1-(1-a)^3600 = 1-(1-2.779x10^-7)^3600 = 1-0.999 = 0.001

`My calculations are therefore self consistent. The time intervals`

`are all treated as independent events. Your interpretation of`

`"moment" is perhaps a conceptual problem.`

Kind regards, MoB

Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/