We are given the following information concerning the three production machines;
![\begin{gathered} Of\text{ the total production;} \\ A=55\text{ \%}=0.55 \\ B=25\text{ \%}=0.25 \\ C=20\text{ \%}=0.20 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lm0hqdldhl681q98pmqg4s8bpydxgmqq2u.png)
Also, we are given the percentage of defective output as follows;
![\begin{gathered} A=3.5\text{ \%}=0.035 \\ B=4.5\text{ \%}=0.045 \\ C=5.5\text{ \%}=0.055 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eoz50e2f07nyrbeieeh3jorwmgh9nxi3zd.png)
Therefore, if an item is selected randomly, the probability that the item is defective would be;
![\begin{gathered} P\lbrack defective\rbrack=(0.55*0.035)+(0.25*0.045)+(0.20*0.055) \\ P\lbrack\text{defective\rbrack}=0.01925+0.01125+0.011 \\ P\lbrack\text{defective\rbrack}=0.0415 \end{gathered}]()
ANSWER:
The probability that the item is defective would be 0.0415