Answer:
Step-by-step explanation:
airplane has four engines
n=4
p = 0.001
This problem is subject to binomial distribution
The plane will crash if 2 or fewer engines are working properly
X - number of engines failed
a) P(crash) = p(x=2 engines fail) + p(x=3 engines fail) + p(x=4 engines fail)
= 4C2 (0.001)^2(0.999)^2 + 4C3 (0.001)^3(0.999)^1 + 4C4 (0.001)^4(0.999)^0
= 0.000005991
[4C2 = 4!/2!(4-2)!]
b) engine 1 will not fail, 3 engines can fail
P(crash) = p(2 engines fail) + p(3 engines fail)
= 3C2 (0.001)^2(0.999)^1 + 3C3 (0.001)^3(0.999)^0
= 0.000002998
c) engine 1 will fail, 3 engines work properly
If even 1 engines fails, plane will crash
P(will not crash) = 3C0 (0.001)^0(0.999)^3 + 3C1 (0.001)^1(0.999)^2
= 0.9999