A company has 2 machines that produce widgets. an older machine produces 23% defective widgets, while the new machine produces o
nly 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?
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.
if a panda eats 20 kg of bamboo per day he eats 140 kg of bamboo every week. multiply the amount he eats per week (140) by the amount of weeks (6) and you get 840 kg of bamboo