Download Probability Concepts

RULES OF PROBABILITY
Complementary Events
If
is the complement of an event A in a sample space S, then
Example: A manufacturing process produces about 2% of items that are
defective. Suppose one item is selected at random from this
manufacturing process. Let us define event A = selected item is defective.
Then P(A) = 0.02. The complement of event A is = selected item is not
defective. Therefore,
, that is, 98% of the
items manufactured are nondefective.
Additive Rule
If A and B are any two events, then
P(A U B) = P(A) + P(B) – P(AB)
If A and B are mutually exclusive, then
P(A U B) = P(A) + P(B)
Example: Suppose there are 50 students enrolled in the calculus course,
45 students enrolled in the machine design course, and 10 in both. Select
one student at random from this group of 100 students.
Let
A = Student is enrolled in a calculus course
B = Student is enrolled in a signal processing course
Find (a) the probability that the student is enrolled in at least one of two
classes, and (b) the probability that the student is enrolled in none of
these classes.
Solution
(a) P(A U B) = P(A) + P(B) – P(A  B)
= 0.50 + 0.45 – 0.10 = 0.85
(b)
= 1 – 0.85 = 0.15
Multiplicative Rule
If A and b are any two events, then
P(AB) = P(A) P(B|A) = P(B) P(A|B)
If A and B are independent, then
P(AB) = P(A)P(B)
Example: Records indicate that for the parts coming out of a hydraulic
repair shop at an airline rework facility, 20% will have a shaft defect,
10% will have a bushing defect, and 75% will be defect-free. For an item
chosen at random from this output, find the probability of the following:
A: The item has at least one type of defect.
B: The item has only a shaft defect.
Solution:
The percentages given imply that 5% of the items have both a shaft
defect and a bushing defect.
Let D1 denote the event that an item has a shaft defect and D2 the event
that it has a bushing defect. Then A = D1 U D2 and
P(A) = P(D1 U D2) = P(D1) + P(D2) – P(D1D2)
= 0.20 + 0.10 – 0.05 = 0.25
Another possible solution is
P(A) = 1 – P
= 1 – 0.75 = 0.25
To find P(B), note that D1 = B U D1D2 where B and D1D2 are mutually
exclusive. Therefore,
P(D1) = P(B) + P(D1D2)
or P(B) = P(D1) – P(D1D2) = 0.20 – 0.05 = 0.15
The Venn Diagram would be
Example: A company manufacturing ballpoint pens has two assembly
lines. A random sample of produced ballpoint pens is taken periodically
and inspected. The inspected pens are classified as defective-trash,
defective-to be fixed, and nondefective. Table 4.19 shows the result of
one such inspection.
Table 4.19 Classification of Manufactured Pens
Assembly line 1 Assembly line 2
Defective-trash
8
2
Defective-to be fixed
13
27
Nondefective
59
91
Total
80
120
Total
10
40
150
200
Define events as follows: A1 = Produced by assembly line 1, A2 =
Produced by assembly line 2, DT = Defective-trash, DF = Defective-to be
fixed, ND = Nondefective
Suppose a pen is selected at random.
(a) What is the overall defective-trash rate?
(b) What is the probability that it is a defective-trash pen
manufactured by assembly line 1?
(c) Suppose a pen selected at random was manufactured by assembly
line 1. Then what is the probability that it is defective-trash?
Solution:
(a) The overall defective-trash rate is 10/200 = .05
(b) P(DT  A1) = 8/200 = 0.04
(c)
= (8/200)/(80/200) = 0.10
Example:
Are passengers wearing seat belts less likely to suffer injuries compared
to those not wearing seatbelts? Data collected by the Maine Dept. of
Public Safety is shown in Table 4.20.
Table 4.20 Seat Belt Usage versus Injury
Injured?
Total
Count
Yes
No
Seat Belt?
Yes
3,984
57,107
61,091
No
2,323
8,138
10,461
Total
6,307
65,245
71,552
Suppose a passenger is selected at random. Find (a) the probability that a
passenger wearing a seatbelt is injured, and (b) the probability that a
passenger not wearing a seatbelt is injured.
Solution:
P(Injury|seat belt) = 3,984/61,091 = 0.065 and
P(Injurt|No seat belt) = 2,323/10,461 = 0.222
Bayes’ Rule
If B1, B2, …, Bk form a partition of S, and if A is any event in S then
Example:
A company buys tires from two suppliers, 1 and 2. Supplier 1 has a record
of delivering tires containing 10% defectives, whereas supplier 2 has a
defective rate of only 5%. Suppose 40% of the current supply came from
supplier 1. If a tire is selected randomly from this supply and observed to
be defective, find the probability that it came from supplier 1.
Solution:
Let
denote the event that a tire comes from supplier i(i = 1,2), and
note that B1 and B2 form a partition of the sample space for the
experiment of selecting one tire. Let A denote the event that the selected
tire is defective. Then