Mathematics
skittle416
12

A company has 2 machines that produce widgets. an older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. in addition, the new machine produces 3 times as many widgets as the older machine does. given that a widget was produced by the new machine, what is the probability it is not defective?

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(1) Answers
leare01

Events A = widget was produced by the old machine B = widget was produced by the new machine. D = widget was defective Recall definition of conditional probability P(X|Y)=P(X n Y)/P(Y)    [ n represents set intersection operator ] By the law of total probability, P(A)=1/(1+3)=1/4 P(B)=3/(1+3)=3/4 Given: Defective rate of new machine P(D|B)=P(D n B)/P(B)=0.08  => P(D n B)=0.08*0.75=0.0600 Since P(B)=P(D n B)+P(~D n B)=0.75, this means that P(~D n B)=0.75-0.0600 = 0.69 Proceed to calculate probability of non-defective widget given it is produced by the new machine: P(~D|B) = P(~D n B)/P(B) = 0.69 / 0.75 = 0.92 Conclusion The probability that the widget is not defective given that it was produced by the new machine is 0.92.

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