In a batch of 840 calculators, 10 were found to be defective. What is the probability that a calculator chosen at random will be defective? Write the probability as a percent. Round to the nearest tenth of a percent if necessary.

10 pieces in 840 calculators are found to be defective...that means that in every 84 pieces.. there is one defect piece.... 1 calculator is chosen at random... then the probability that you'll choose at random and find a defective is #1/84#
Divide
#color(white)(84||)0.0119#
#84bar(| color(white)(d)1color(white)(dddd)#
#color(white)(||color(white)(i)) darr color(white)(dddd)#
#color(white)(dd)bar(color(white)(d)10color(white)(ddd)#
#color(white)(||color(white)(i)) darr color(white)(dddd)#
#color(white)(dd)bar(color(white)(d)100color(white)(ddd)#
#color(white)(ddiii)84#
#color(white)(dd)bar(color(white)(dii)16color(white)(ddd)#
#color(white)(||color(white)(iii)) darr color(white)(dddd)#
#color(white)(dd)bar(color(white)(dii)160color(white)(ddd)#
#color(white)(ddiiiii)84#
#color(white)(dd)bar(color(white)(iiiii)76color(white)(ddd)#
#color(white)(||color(white)(iiii)) darr color(white)(dddd)#
#color(white)(dd)bar(color(white)(iiiii)760color(white)(ddd)#
#color(white)(ddiiiii)756#
#color(white)(dd)bar(color(white)(iiiiiiii)4color(white)(ddd)#

It's gonna go on an on so I'm gonna stop right there.. you get
#1/84~~0.0119#

Now on the part to make it a percent
You know that #x.yz=(xyz)/100#(here these variables don't multiply they are just like a single number)
So... #0.0119=119/10000#
Which is approximately
#1.19/100#
So it is #1.19%#
It is approximately #1%# rounded