An airplane has 4 engines. If each of an airplane's engines has a 0.01 probability of stalling and the probability that an airplane will fly safely is 0.9 if all engines are running, 0.7 if only 3 work, and 0.3 if less than 3 engines work, calculate the expectation that the plane has to crash.
1st scenario, 4 engines fail
The probability that the 4 engines stop is $0.01*0.01*0.01*0.01 = \frac{1}{10^{-8}}$
If 4 engines don't work the probability that the plane has to crash is 1 - 0.3 = 0.7 so scenario 1 has a probability $\frac{1}{10^{-8}}* 0.7$ to happen.
2nd scenario, 3 engines fail
The probability of 3 engines stopping is $0.01*0.01*0.01*0.99 = 0.99*\frac{1}{10^{-6}}$
If 3 engines don't work the probability that the plane has to crash is 1 - 0.3 = 0.7 so scenario 2 has $\frac{1}{10^{-6}}* probability of happening.
3rd scenario, 2 engines fail
The probability of 2 engines stopping is $0.01*0.01*0.99*0.99 = 0.98*\frac{1}{10^{-4}}$
If 2 engines don't work the probability the plane has to crash is 1 - 0.3 = 0.7 so scenario 3 has a probability $\frac{1}{10^{-4}}* 0.69 $ to happen.
4th scenario, 1 engine fails
The probability of 1 engines stopping is $0.01*0.99*0.99*0.99 = 0.97*\frac{1}{10^{-2}}$
If 1 engines don't work the probability that the plane has to crash is 1 - 0.7 = 0.3 so scenario 4 has a probability $\frac{1}{10^{-2}}* 0.29 $ to happen.
5th scenario, 0 engines fail
The probability of 0 engines stopping is $1 - 0.99*0.99*0.99*0.99 = 0.04$
If 0 engines don't work the probability of the plane crashing is 1 - 0.9 = 0.1 so scenario 5 has probability $0.1*0.04 = 0.004 $ to happen.
When adding the 5 scenarios we get the answer?
I'm feeling like I've simplified the issue.
Thanks for any help.