Answer: 0.07938
Step-by-step explanation:
Let X be a binomial variable that represents the components of a specific engine.
As per given , the probability of componentis found to not perform to standards : p= 1%=0.01
Binomial probability formula :
[tex]P(X=x)=\ ^nC_xp^x(1-p)^{n-x}[/tex]
Sample size : n= 100
If more than two are found to be non-performing the entire shipment is returned.
Now, Required probability : [tex]P(X>2)=1-P(X\leq 2)[/tex]
[tex]=1-(P(X=0)+P(X=1)+P(X=2))\\\\=1- (^{100}C_{0}(0.01)^0(0.99)^{100}+^{100}C_{1}(0.01)^{1}(0.99)^{99}+^{100}C_{2}(0.01)^{2}(0.99)^{98})\\\\=1-((0.99)^{100}+(100)(0.01)(0.99)^{99}+\dfrac{100!}{2!98!}(0.01)(0.99)^98)\\\\=1-(0.36603+0.36973+0.18486)\\\\=1-(0.92062)\\\\=0.07938[/tex]
So, the probability that a shipment is returned = 0.07938