Asked by koro
The company will use machine A,B,C,D with probabilities 0.23,0.45,0.19 and 0.13 respectively.From past experience it is known that the probability of defective tablet produced by the machine are 0.11,0.2,0.07 and 0.05 respectively.Suppose a defective tablet is produced： a)what is the probability of machineD? b)what is the probability of machineB?
Answers
Answered by
gothic vampire
Apply Bay's rule here
A-p(E4/D)=)0.05*0.13/(0.23*0.11)+(0.45*0.2)+(0.19*0.07)+(0.13*0.05)
=5/196
A-p(E4/D)=)0.05*0.13/(0.23*0.11)+(0.45*0.2)+(0.19*0.07)+(0.13*0.05)
=5/196
Answered by
gothic vampire
Similarly you can find for machine B
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