What is the probability that 15 witnesses would die unnaturally within 1yr of the JFK assassination?

[Richard Charnin]

26. Have you actually read the math proof?

Apparently, you either have not read it or fail to comprehend it.

I will repeat the basic problem for you.
Its really very simple.

Given:
a group of X individuals

Determine:
the probability that at least n of the X in the group would die UNNATURAL deaths in the same year.

Problem:
Determine the probability that any individual would an UNNATURAL death (i.e. gunshot, karate chop, airplane crash, automobile crash, etc.) in any given year.

Probability of an UNNATURAL death in 1 year
suicide….. 0.000107
homicide…. 0.000062
accidental.. 0.000359
undetermined 0.000014

Total probability= 0.000542

Solution:
You have a group of X =1400 persons
p = .000542 = PROBABILITY of of an UNNATURAL death in any given year.

Therefore, the expected number (a) of UNNATURAL deaths in a group of 1400 is equal to the probability of an unnatural death times the number of witnesses:

a = 0.7588 = p*N = 000542*1400
In other words, we can expect that 0.7588 (one) person out of 1400 would die UNNATURALLY IN ANY GIVEN YEAR.

The probability that m=15 of 1400 in a group would die UNNATURALLY in a given year is calculated using the Poisson probability function:
P (m) = a^m * exp (-a) / m!

Plugging into the formula:
P(15) = 0.7588^15* exp (-.7588)/15!

Now, you do the math.

Note: The London Times failed to analyze the probability of UNNATURAL DEATHS.
They did not correctly specify the problem.

It is not the number of deaths that needs to be modeled.
It's the number of UNNATURAL deaths that must be considered.

Ok, now do the math... or forever stick with the no-conspiracy theory that Oswald, the lone nut gunman, did it. But if Lee Harvey did it, then you would not have all those UNNATURAL DEATHS, now would you?