Answer:
0.762
Step-by-step explanation:
We are given that
Factory A produces 4 times as many computers as factory B
Let [tex]E_1 and E_2[/tex] be the event denotes computers produced by factory by A and factory by B.
Let [tex]P(E_2)=x [/tex]
Then[tex] P(E_1)=4x[/tex]
Sum of all probabilities=1
[tex]x+4x=1[/tex]
[tex]5x=1[/tex]
[tex]x=\frac{1}{5}[/tex]
Hence, [tex]P(E_1)=\frac{4}{5},P(E_2)=\frac{1}{5][/tex]
Let A denotes the event that an item is defective.
[tex]P(A/E_1)=0.034,P(A/E_2)=0.034[/tex]
We have to calculate [tex]P(E_1/A)[/tex]
[tex]P(E_1/A)=\frac{P(E_1)\cdot P(A/E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)\cdot P(A/E_2)}[/tex]
[tex]P(E_1/A)=\frac{\frac{4}{5}\times 0.034}{\frac{4}{5}\times 0.034+\frac{1}{5}\times 0.034}[/tex]
[tex]P(E_1/A)=\frac{0.0272}{0.0357}[/tex]
[tex]P(E_1/A)=0.762[/tex]
Hence, the probability that the item produced by factory A and it is found to be defective=0.762