+117 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
+129852 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
