Thread: probability

Originally Posted by alderon

box (A) contains 8 items of w/c 3 are defective
box (B) contains 5 items of w/c 2 are defective.
an item is drown from each box,if one item is defective and one is not.
what is the probability that the defective item came from box (A)?

Let p(Ad) denote the probability that item drawn from A is defective, and
P(Bd) the corresponding probbaility for the item drawn from B.

Then Bayes theorem tells us:

p(Ad| Ad or Bd and not(Ad and Bd))= p(Ad or Bd and not(Ad and Bd)|Ad)p(Ad)/ p(Ad or Bd and not(Ad and Bd))

Now:

p(Ad)=3/8,

p(Ad or Bd and not(Ad and Bd))=3/8+2/5-(3/8)(2/5),

p(Ad or Bd and not(Ad and Bd)|Ad)=1-(2/5)

RonL

i got it...