Suppose that ten items are chosen at random from a large batch delivered to a company. The manufacturer claims that just 8% of the items in the batch are defective. Assume that the batc
enough so that even though the selection is made without replacement, the number 0.08 can be used to approximate the probability that any one of the ten items is defective. In addition, as
that because the items are chosen at random, the outcomes of the choices are mutually independent. Finally, assume that the manufacturer's claim is correct. (Round your answers to
three decimal places.)
(a) What is the probability (as a %) that none of the ten are defective?
Represent each selection of ten items as a string of ten letters, each either D (for defective) or N (for not defective). Thus, NNNNNNNNNN represents a selection with no defective items.
assumption, the choices for each item are dependent, and the probability that any individual item is not defective is approximately 0.92. Thus,
P(NNNNNNNNNN) = (0.92) (0.92) (0.92) (0.92) (0.92) (0.92) (0.92) (0.92) (0.92) (
%.
(b) What is the probability (as a %) that at least one of the ten is defective?
%
(c) What is the probability (as a %) that exactly three of the ten are defective?
A selection with three defective items can be obtained in many different ways: DNNNDDNNNN, NNDDDNNNN, NNDNDNNNDN, and so forth. For example, by similar reasoning as in part (a), t
probability of selecting NNNDNNDNDN is
P(NNNDNNDNDN) = (0.92)$^3$(0.08)$^1$(0.92)$^1$(0.08)$^1$(0.92)$^1$(0.08)$^1$(0.92)$^1$
(0.92) (0.08).
Now there are as many different selections with 3 D's as there are ways to choose 3 positions out of a set of 10 positions (for where to place the D's). Thus, the number of different selections
with 3 D's is
It follows by the addition rule that the probability of obtaining 3 defectives is
(0.92) (0.08)
or approximately
%.
