Let X= the number of defective components: X∼Bin(n,p)
P(X=x)=(xn)px(1−p)n−x Given that n=24,p=0.04
The probability that this whole shipment will be accepted is
P(X<3)=P(X=0)+P(X=1)+P(X=2)=
=(024)0.040(1−0.04)24−0+(124)0.041(1−0.04)24−1+
+(224)0.042(1−0.04)24−2≈
≈0.375413+0.375413+0.179886≈0.9307
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