A manufacturer produces soda cans and a quality control worker randomly selects two cans from the assembly line for testing. Past statistics show that 14% of the cans are defective. What is the probability that the two selected cans are defective if the quality control worker selects the two cans from a batch of 50 cans?

P(Both defective) = 7/50

P(Both defective) = 49/2,500

P(Both defective) = 3/175

P(Both defective) = 7/25

failurewithasmile
Apr 15, 2017

#1**+2 **

P ( 2 defective cans from 50) =

14% defective rate means that 7 cans out of 50 should be defective

So.....we have a 7/50 chance of getting a defective can on the first draw and a 6/49 chance of getting a defective can on the second draw.....so

7/50 * 6/49 =

7/49 * 6/50 =

1/7 * 3/25 =

3/175

CPhill
Apr 15, 2017