Solution
Back-up Theory
Let p be the probability a O-ring failing. Then, given
‘two O-rings per booster rocket and that only one required to be functional for the booster rocket it was inside to work’
=> probability of a booster rocket working
= P(at least one O-ring functions)
= 1 - P(neither O-ring functions)
= 1 – p2, [given O-rings are independent] …..…………………………………………………. (1)
Assuming 2 booster rockets in the space shuttle and given both must function for the successful launch of the space shuttle,
P(successful launch) = [1 – p2]2 ……………………………………………………………… (2)
Now to work out the solution,
Part (a)
Here p = 0.3 [given that an individual ring failed 30% of the time]. Then vide (2),
P(successful launch) = [1 – 0.32]2
= 0.8372
=> P(catastrophic failure)
= 1 – 0.8372
= 0.1628 Answer
Part (b)
Here p = 0.66
P(successful launch) = [1 – 0.662]2
= 0.3185
=> P(failure or explosion)
= 1 – 0.3185
= 0.6815 Answer
Part (c)
We want P(failure) < 0.05
=> P(successful launch) > 1 – 0.05 = 0.95
i.e., [1 – p2]2 > 0.95 [vide (2)]
or 1 – p2 > 0.9950
=> p2 < 0.0050
Or, p < 0.0707
Thus, the probability of a single O-ring failing should be less than 7% Answer
DONE
Problem 3 (Systems Engineering) The Challenger was a space shuttle that exploded on January 28, 1986...
Problem 5. Each space shuttle has siz 0-rings and each O-ring fails with probability ea+bt where a 5.085, b--0.1156, and t is the temperature (in degrees Fahrenheit) at the time of the launch of the space shuttle. At the time of the fatal launch of the Challenger, t 31 (a) Let X be the number of failing O-rings at launch temperature 31.F. What type of probability distribution does X have, and what are the values of its parameters? b) What...