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Sagot :
We are given the following information concerning the three production machines;
[tex]\begin{gathered} Of\text{ the total production;} \\ A=55\text{ \%}=0.55 \\ B=25\text{ \%}=0.25 \\ C=20\text{ \%}=0.20 \end{gathered}[/tex]Also, we are given the percentage of defective output as follows;
[tex]\begin{gathered} A=3.5\text{ \%}=0.035 \\ B=4.5\text{ \%}=0.045 \\ C=5.5\text{ \%}=0.055 \end{gathered}[/tex]Therefore, if an item is selected randomly, the probability that the item is defective would be;
[tex]\begin{gathered} P\lbrack defective\rbrack=(0.55\times0.035)+(0.25\times0.045)+(0.20\times0.055) \\ P\lbrack\text{defective\rbrack}=0.01925+0.01125+0.011 \\ P\lbrack\text{defective\rbrack}=0.0415 \end{gathered}[/tex]ANSWER:
The probability that the item is defective would be 0.0415
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