Question 1143700
.
<pre>
(1)  The probability to randomly select one satisfactory piston from Machine 1 production is  {{{392/(392 + 98)}}} = {{{392/490}}} = {{{4/5}}}.


(2)  The probability to randomly select one unsatisfactory piston from Machine 2 production is  {{{120/(480 + 120)}}} = {{{120/600}}} = {{{1/5}}}.


(3)  Notice that the production sets from Machine 1 and from Machine 2 are <U>disjoint</U> sets.


     It implies that the final probability under the question is the product of particular probabilities


         P = {{{4/5}}}.{{{1/5}}} = {{{4/25}}} = 0.16 = 16%.    <U>ANSWER</U>
</pre>