Answer to Question #193610 in Statistics and Probability for Kierra

Question

A factory produces pistons for cars machine one produces 287 satisfactory pistons and 123 unsatisfactory pistons machine to produces 208 satisfactory pistons and 192 unsatisfactory pistons what is the probability that the Pistons chosen for machine one is unsatisfactory and the piston chosen from machine to a satisfactory

Answer 1

$n\left(Total\:machine\:1\right)=287+123$

$\:n\left(Unsatisfactory\:machine\:1\right)=123\:\:\:\:$

$P\left(Unsatisfactory\:machine\:1\right)=\frac{n\left(Unsatisfactory\:machine\:1\right)}{n\left(Total\:machine\:1\right)}$

$P\left(Unsatisfactory\:machine\:1\right)=\frac{123}{410}\:=0.3$

$n\left(Total\:machine\:2\right)=240+60=300$

$\:n\left(Satisfactory\:machine\:1\right)=240\:\:$

$P\left(Satisfactory\:machine\:2\right)=\frac{n\left(Satisfactory\:machine\:2\right)}{n\left(Total\:machine\:2\right)}$

$P\left(Satisfactory\:machine\:2\right)=\frac{240}{300}=0.8$

$P\left(Unsatisfactory\:machine\:1\cap Unsatisfactory\:machine\:2\right)=$

$P\left(Unsatisfactory\:machine\:1\right)\cdot P\left(Unsatisfactory\:machine\:2\right)$

$\:=0.3\times 0.8=0.24$