combination of normal and binomial distributions.
Calculate probability of opening below 100m.
mean=175 m
standard deviation = 35 m
z=(100-175)/35=-15/7
p=P(Z<z)=0.01606229 from normal probability tables.
Calculate probability of at least one parachute opens below 100m
p=0.01606229n=5
x=0
P(X>0)
=1-P(x=0)
=1-C(5,0)p^x(1-p)^5
=1-1*0.01606229^0*(1-0.01606229)^5
=1-0.9222 (approximately)
=0.0778
Answer:
Probability that at least one of five cargoes will be damaged is 0.0778