Question

A factory produces widgets. At the end of production, there is a quality assurance process that rejects all faulty widgets. On the first day of the month, the factory produced 211 widgets, 13 of which were faulty. On the second day, the factory produced 188 widgets, 20 of which were faulty. On the third day, the factory produced 201 widgets, 26 of which were faulty. Based on these 3 days, what is the experimental probability that a unit is not faulty?

A.13/600
B.59/600
C.541/600
D.59/60

Answer 1

To solve for the problem above we must first solve the probability of the non- faulty widgets from the faulty widgets and define what each given data are. The formula for the experimental probability is p(a) = the number of the not faulty widgets / the number of widgets, where a is the number of days.
the number of not faulty widgets in day 1 is 211 - 13 = 198, day 2 is 188 - 20 = 168 ; day 3 is 201 - 26 = 175. To sum it up
p(a) = (198 + 168 + 175) / (211 + 188 + 201) = 541/600.
The answer is letter C.

Answer 2

The factory produced a total of 600 widgets for three days of operation. During these days, the factory has produced 59 faulty widgets. This means that there are 541 widgets that are not faulty.

To calculate the experimental probability that the unit is not faulty, divide the number of not faulty widgets by the total number.

The division gives 541/600. Thus, the answer is letter C.