Question

A bicycle manufacturing plant is producing the first few frames for their new electric bicycle design. The first batch of 60 frames has some issues: 1/3 of the frames had a weak junction where top tube meets the seat tube. After some fine tuning, a second batch of 120 frames is made, of which 1/4 are bad, still having the weak junction. Finally, another bug is fixed and a third batch of 180 frames is made and only 1/5 of these are bad. Unfortunately, due to a clerical error, all 360 frames were combined, unlabeled, into the same shipping bin. You have access to a quick test that checks if a bike frame is defective, which is correct with probability 9/10. That is, if the bike frame is defective, the test will say it is defective with probability 9/10. If the bike frame is not defective, the test will say it is defective with probability 1/10.

A) Determine the probability that the test returns "defective," P[D].B) Determine the probability that, given the test returns "not defective," the bike frame is not bad P[Bc |Dc].

Answer 1

a) 131/450

b) 1233/1276

Explanation:

P(bad) = P(1st batch)*P(bad 1st batch ) + P(2nd batch )*P(bad 2nd batch) + P(3rd batch )*P(bad 3rd batch)

p(bad) =(60/360)*(1/3) + (120/360)*(1/4 ) + (180/360)*(1/5)

= 43/180

And that of P(good )

= 1 - 43/180

= 137/180

a)

P(defective) = P(bad)*P(defective /bad) + P(good)*P(defective /good)

= (43/180)*(9/10) + (137/180)*(1/10)

= 131/450

b)

P(Bc I Dc ) = P(good)*P(not defective |good) / P(not defective)

= (137/180)*(1 - 1/10) / (1 - 131/450)

= 1233/1276