9/11 Terror Attack Probability

By Bayes’s Theorem
http://en.m.wikipedia.org/wiki/Bayes’_theorem

1. First WTC (World Trade Center) Attack:

Prior Probability
Initial estimate that terrorists would crash planes into Manhattan skyscrapers = 1/20,000 = x = 0.005%

New Event: 1st Plane hits WTC
Probability of plane hitting if terrorists are attacking Manhattan skyscrapers = y = 100%

Probability of plane hitting if terrorists are NOT attacking Manhattan skyscrapers (ie accident) = z = 2 cases in last 25,000 days (once in 1945, another in 1946) = 1/12,500 = 0.008%

Posterior Probability
Revised estimate of probability of terror attack, given first plane hitting WTC =
\displaystyle\boxed{ \frac{xy}{xy+z(1-x)} = 38\%}

2. Second WTC Attack

Now 38% becomes the prior probability = x
y, z remain the same.

Reapply the Bayes’s Theorem:

Posterior Probability
Revised estimate of probability of terror attack, given 2nd plane hitting WTC =
\displaystyle\boxed{ \frac{xy}{xy+z(1-x)} = 99.99\%}

There was a 18-minute interval between the first and the second attack. If someone who had calculated the Bayesian Math to predict 99.99% chance of the would-be second attack, the entire people on the second building could have been evacuated right after the first tower attack, instead of being advised naively (0.01% chance it was safe) to return to their office in the second tower !

You can say this is déjà-vu, but it is Math which could have saved thousands of life on that tragic day of 9/11.

Watch this disturbing documentary video of the 9/11 Attack:

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One thought on “9/11 Terror Attack Probability

  1. Pingback: Baeysian Probability Could Help Search MH370 Missing Plane | Math Online Tom Circle

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