Let X= the number of defective components: X∼Bin(n,p).
Given n=10,p=0.1,q=1−p=1−0.1=0.9.
i)
P(X<2)=P(X=0)+P(X=1)
=(010)(0.1)0(0.9)10−0+(110)(0.1)1(0.9)10−1
=0.7360989291 The probability that the consignment is accepted is 0.7360989291.
ii)
P(X≥2)=1−P(X<2)
=1−(010)(0.1)0(0.9)10−0−(110)(0.1)1(0.9)10−1
=0.2639010709
The probability that the consignment is rejected is 0.2639010709.
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