p = probability defective = 6/16 = .375
1 - p = 10/16 = .625 = probability good
we do three trials on the box. They all must be good if it is to pass
probability of first trial fuse good = 10/16
probability of second choice being good fuse = 9/15
third = 8/14
so probability of passing = (10*9*8)/(16*15*14) = .0762
so probability of rejection = 1-.0762 = .924
A factory produces fuses, which are packaged in boxes of 16. Three fuses are selected at random from each box for inspection. The box is rejected if at least one of the three fuses is defective. What is the probability that a box containing six defective fuses will be rejected?
3 answers
arithmetic!
(10*9*8)/(16*15*14) = .2143
1 - .2143 = .7857
(10*9*8)/(16*15*14) = .2143
1 - .2143 = .7857
the probability of at least one being defective equals
1 minus the probability that all three are good
1 - (5/8)^3 = .756 (approx)
1 minus the probability that all three are good
1 - (5/8)^3 = .756 (approx)