Answer:
0.226
Step-by-step explanation:
Given that:
No wing crack (x) = 70%
Detectable wing crack (y) = 25%
Critical wing crack (z) = 5%
Number of samples (n) = 5
Probability that atleast a plane has critical crack:
1 - p(no critical crack)
Using the multinomial distribution calculator :
P(5,0,0) + p(4,1,0) + p(3,2, 0) + p(2,3,0) + p(1,4,0) + p(0,5,0)
From the calculator :
P(5,0,0) + p(4,1,0) + p(3,2, 0) + p(2,3,0) + p(1,4,0) + p(0,5,0) = 0.7741
1 - 0.7741 = 0.2259
= 0.226
