Question 1056997: 5.92 The Challenger Disaster. In a letter to the editor that appeared in the February 23, 1987, issue of U.S. News and World Report, a reader discussed the issue of space-shuttle safety. Each “criticality 1” item must have a 99.99% reliability, by NASA standards, which means that the probability of failure for a “criticality1” item is only 0.0001. Mission 25, the mission in which the Challenger exploded on takeoff, had 748 “criticality 1” items.
Use the Poisson approximation to the binomial distribution to determine the approximate probability that
a. none of the “criticality 1” items would fail.
b. at least one “criticality 1” item would fail.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! n must be greater than or = 100, our n is 748
Probability(Pr) must be less than or = 0.01, our Pr is 0.0001
:
our constant lambda is 748 * 0.0001 = 0.0748
:
a) Pr ( x = 0 ) = (e^(-0.0748) * 0.0748^0) / 0! = 0.9279
:
b) Pr ( x > or = 1 ) = 1 - Pr ( x = 0 ) = 1 - 0.9279 = 0.0721
:
Note that Dr Richard Feynman proved that by NASA's ignoring the advice of Morton Thiercol rocket engineers and the physics of cold on rubber seals they created a 100% chance of failure
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