10 A longtime supplier of mechanical ents installed in proje

10. A long-time supplier of mechanical ents installed in projects your company constructs has a history of shipping you components which may exhibit one or both categories of defects: orflaws in the finish (minor defects, but may be setting to your customer) and mechanical defects (major defects, component must be replaced). The number of each type defect found in an individual component has been recorded and is a random variable. Company records for the most components received have resulted in the joint probability recent 5 distribution shown in the following table, where x represents the occurrence of finish defects and Y represents the occurrence of mechanical defects. o 11 50 4 50 2750 1750 1/50 1/50 1 8 50 3 50 2750 1/50 1/50 2 3 50 2 50 1 50 4 50 a 3/ 50 1/50 4 1/50

Solution

a) Marginal Probability distribution

Y(0) = P(0,0)+P(1,0)+P(2,0)+P(3,0)+P(4,0)+P(5,0)=(11/50)+(4/50)+(2/50)+(1/50)+(1/50)+(1/50)= 2/5

Y(1)=P(0,1)+P(1,1)+P(2,1)+P(3,1)+P(4,1)=(8/50)+(3/50)+(2/50)+(1/50)+(1/50)=15/50

Y(2)=P(0,2)+P(1,2)+P(2,2)+P(3,2)=(4/50)+(3/50)+(2/50)+(1/50)= 1/5

Y(3)=P(0,3)+P(1,3)=(3/50)+(1/50)=2/25

Y(4)=P(0,4)=1/50

b) P(0,Y)=P(0,0)+P(0,1)+P(0,2)+P(0,3)+P(0,4)=(11/50)+(8/50)+(4/50)+(3/50)+(1/50)= 27/50

c) F(0,0)=11/50

F(0,Y)=27/50

F(X,0)=20/50

since F(0,0) is not equal to [ F(0,Y)*F(X,0)]

Therefore it is dependent.