Page 64 - Applied statistics and probability for engineers

42 Chapter 2/Probability

Surface flaw
360 No Yes 40
400 400

Defective
342 No Yes 18 30 No Yes 10
360 360 40 40

FIGURE 2-14 Tree diagram for parts classified.

Random Samples and Conditional Probability
Recall that to select one item randomly from a batch implies that each item is equally likely to
be picked. If more than one item is selected, randomly implies that each element of the sample
space is equally likely to be picked. When sample spaces were presented earlier in this chap-

ter, sampling with and without replacement was deined and illustrated for the simple case of
a batch with three items { , , }a b c . If two items are selected randomly from this batch without
{
replacement, each of the six outcomes in the ordered sample space ab,ac,ba,bc,ca,cb} has
probability 1 6. If the unordered sample space is used, each of the three outcomes in {{ , },a b
/
c
a
b
c
{ , },{ , }} has probability 1 3/ .
When a sample is selected randomly from a large batch, it is usually easier to avoid enu-
meration of the sample space and calculate probabilities from conditional probabilities. For
example, suppose that a batch contains 10 parts from tool 1 and 40 parts from tool 2. If two
parts are selected randomly, without replacement, what is the conditional probability that a
part from tool 2 is selected second given that a part from tool 1 is selected i rst?
Although the answer can be determined from counts of outcomes, this type of question can
be answered more easily with the following result.
Random Samples
To select randomly implies that at each step of the sample, the items that remain in
the batch are equally likely to be selected.

If a part from tool 1 were selected with the irst pick, 49 items would remain, 9 from tool 1 and

40 from tool 2, and they would be equally likely to be picked. Therefore, the probability that a
part from tool 2 would be selected with the second pick given this irst pick is

(
P E E 1) = 40 49
2 |
/
In this manner, other probabilities can also be simplii ed. For example, let the event E
consist of the outcomes with the i rst selected part from tool 1 and the second part from tool
2. To determine the probability of E, consider each step. The probability that a part from
tool 1 is selected with the i rst pick is P E 1 ( ) = 10 50. The conditional probability that a part
/
from tool 2 is selected with the second pick, given that a part from tool 1 is selected i rst, is
(
P E E 1) = 40 49. Therefore,
2 |
/
P E ( ) = ( 2 | ) 40 10 = 8
.
P E E P E ( ) =
1
1
49 50 49
Sometimes a partition of the question into successive picks is an easier method to solve the problem.
Example 2-24 Random Inspection Consider the inspection described in Example 2-14. Six parts are selected ran-
domly without replacement from a bin of 50 parts. The bin contains 3 defective parts and 47 nondefec-
tive parts. What is the probability that the second part is defective given that the irst part is defective?