Intersection of Events & Multiplication Rule · Multiplication rule Sometimes we may need to find the probability of two events happening together. The probability of the intersection

Intersection of events A and B

Intersection of Events & Multiplication Rule

Intersection of events

Let A and B be two events defined in a sample space. The intersection of A and B represents the collection of all outcomes that are common to both A and B is denoted by any of the followings

A and B, A ∩ B, or AB

Example: Let A = the event that a person owns a PC Let B = the event that a person owns a mobile

A B A and B

42


Multiplication rule

Sometimes we may need to find the probability of two events happening together.

The probability of the intersection of two events A and B is called their joint probability. It is written

P(AB) or P(A ∩ B)

It can be obtained by multiplying the marginal probability of one event with the conditional probability of the second one.

Multiplication rule: The probability of the intersection of two events A and B is

P(AB) = P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)

43


Example: The following table gives the classification of all employees of a company by gender and college degree.

If one employee is selected at random, what is the probability that the employee is a female and a college graduate?

Solution

P(FG) = P(F) P(G|F) = P(G) P(F|G) = P(GF)

Total

Not a college graduate

(N)

College

graduate (G)

27 20 7 Male (M)

13 9 4 Female (F)

40 29 11 Total

9 7 4

F G

44


P(F) = 13/40

P(G|F) = 4/13

P(FG) = P(F) P(G|F) = 4/40

OR

P(G) = 11/40

P(F|G) = 4/11

P(GF) = P(G) P(F|G) = 4/40

Total

Not a college

graduate

(N)

College

graduate

(G)

27 20 7 Male (M)

13 9 4 Female (F)

40 29 11 Total

M

27/40

F

13/40

G|M

7/27

N|M

20/27

G|F

4/13

N|F

9/13

P(MG) = (27/40)(7/27) = 7/40

P(MN) = (27/40)(20/27) = 20/24

P(FG) = (13/40)(4/13) = 4/40

P(FN) = (13/40)(9/13) = 9/40

9 7 4

F G

45


Example: A box contains 20 DVD, 4 of which are defective. If two DVDs are selected at random (without replacement), what is the probability that both are defective?

Solution: Let

G1 = 1st DVD selected is good, G2 = 2nd DVD selected is good

D1 = 1st DVD selected is defective, D2 = 2nd DVD selected is defective

We need to find the joint probability P(D1 D2)

P(D1 D2) = P(D1)P(D2 |D1)

There are 4 defective DVDs P(D1) = 4/20

Given that the 1st DVD selected (without replacement) is defective, there

are 19 DVDs left in which 3 are defective and 16 are good.

P(D2 |D1) = 3/19 P(D1 D2) =(4/20)(3/19) =.0316

46


If events A and B are independent, their joint probability simplifies from

P(A ∩ B) = P(A)P(B |A) TO P(A ∩ B) = P(A)P(B )

Sometimes we know the joint probability of two events A and B, in this case, the conditional probability of B given A or A given B is

given that P( ) 0 and P( ) 0A B

( )( | )

( )

P A BP A B

P B

( )( | )

( )

P A BP B A

P A

47


Example: According to a survey, 60% of all homeowners owe money on home mortgages. 36% owe money on both home mortgages and car loans. Find the conditional probability that a homeowner selected at random owes money on a car loan given that he owes money on a home mortgage.

Solution: Let

H = homeowner owes money on a home mortgage

C = homeowner owes money on a car loan

From the given information

P(H) = .6 and P(HC) = .36

Hence,

P(C|H) = P(CH)/P(H) = .36/.6 = .6

48


Example: A computer company has two quality control inspectors, Mr. Smith and Mr. Robertson, who independently inspect each computer before it is shipped to a client. The probability that Mr. Smith fails to detect a defective PC is .02 while it is .01 for Mr. Robertson. Find the probability that both inspectors will fail to detect a defective PC.

Solution: Let

A = 1st inspector fails to detect a defective PC

B = 2nd inspector fails to detect a defective PC

From the given information

P(A) = .02 and P(B) = .01

Hence,

P(AB) = P(A)P(B) = 0.0002

49


Example: The probability that a patient is allergic to Penicillin is .2. Suppose this drug is administrated to three patients. Find a) The probability that all three of them are allergic to it b) At least one of them is not allergic

Solution: Let

A = 1st patient is allergic to penicillin

B = 2nd patient is allergic to penicillin

C = 3rd patient is allergic to penicillin

50

Union of events

Let A and B be two events defined in a sample space S. The union of events A and B is the collection of all outcomes that belong either to A and B or to both A and B and is denoted by “ A or B” or “A ∪ B”

A B

S

51

Union of Events & Addition Rule

Example: A company has 1000 employees. Of them, 400 are females and 740 are labor union members. Of the 400 females, 250 are union members. Describe the union of events “female” and “union member”

Solution 25

0

150 490

Total Not Union Members Union Members

600 110 490 M

400 150 250 F

1000 260 740 Total

Total employees that are “females” or “union members” or both “females and

union members” are

400 + 740 – 250 = 890

52


Addition rule

The probability of the union of two events A and B is

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Example: A university president has proposed that all students must take a course in ethics as a requirement for graduation. Three hundred faculty members and students from this university were asked about their opinion on this issue and it shown in the following table.

Find the probability that a person selected is a faculty member or in favor.

Opinion

Total Neutral Oppose Favor

70 10 15 45 Faculty

230 30 110 90 Student

300 40 125 135 Total

53


Solution:

P(F ∪ A) = P(F) + P(A) – P(F ∩ A)

P(F ∩ A) = P(F)P(A |F)

P(F) = 70/300 = 0.233 P(A) = 135/300 = 0.45 P(A|F) = 45/70=0.643

P(F ∩ A) = 0.233 x 0.643 = 0.150 or P(F ∩ A) = 45/300 = 0.15

P(F ∪ A) = 0.233 + 0.45 – 0.15 = 0.533

Opinion

Total Neutral (C) Oppose (B) Favor (A)

70 10 15 45 Faculty (F)

230 30 110 90 Student (S)

300 40 125 135 Total

54


Example: There are a total of 7225 thousand persons with multiple jobs in the US. Of them, 4115 thousand are male, 1742 thousand are single, and 905 thousand are male and single. What is the probability that a selected person is a male or single?

Solution: Let us define the following events

M = the selected person is male, A = the selected person is single

P(M) = 4115/ 7225 = 0.57 P(M ∪ A) = P(M) + P(A) – P(M ∩ A)

P(A) = 1742/7225 = 0.241 P(M ∪ A) = 0.57+0.241– 0.125 = 0.686

P(M ∩ A) = 905/7225 = 0.125

Total Married (B) Single (A)

4115 3210 905 Male (M)

3110 2273 837 Female (F)

7225 5483 1742 Total

55


The probability of the union of two mutually exclusive events A and B is

P(A ∪ B) = P(A) + P(B)

Example: A university president has proposed that all students must take a course in ethics as a requirement for graduation. Three hundred faculty members and students from this university were asked about their opinion on this issue and it shown in the following table.

Find the probability that a person selected is in favor or is neutral

56


Opinion

Total Neutral Oppose Favor

70 10 15 45 Faculty

230 30 110 90 Student

300 40 125 135 Total

57


Solution:

Example: Eighteen percent of the working lawyers in the United States are female. Two lawyers are selected at random and it is observed whether they are male or female. a) Draw a tree diagram for this experiment b) Find the probability that at least one of the two lawyers is a female.

Solution:

FF=0.18x0.18 = 0.032

FM=0.18x0.82 = 0.148

MM=0.82x0.82 = 0.672

MF=0.82x0.18 = 0.148

P(at least 1 F) = 1 - 0.672 = 0.328

58


Law of Total Probability & Bayes’ Rule

Consider the following Venn diagram

Can we find the area (probability) of A assuming that we know the probability of each Bi & P(A|Bi) ?

B1 B2 Bn-1 Bn

B

A

A1 A2 An-1 An

Addition Law

Multiplication Law

Conditional Probability

( ) ( ) ( ) ( )P A B P A P B P A B

( ) ( ) ( | )P A B P B P A B

( ) ( ) ( | )P A B P A P B A

( )( | )

( )

P A BP A B

P B

( )( | )

( )

P A BP B A

P A

60


1 1

1

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | )

P A P B AP A B

P A P B A P A P B A

Bayes’ theorem (two-event case)

Bayes’ theorem

2 22

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | )

P A P B AP A B

P A P B A P A P B A

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | ) ... ( ) ( | )

i ii

n n

P A P B AP A B

P A P B A P A P B A P A P B A

61


Example: Manufacturing firm that receives shipment of parts from two different suppliers. Currently, 65 percent of the parts purchased by the company are from supplier 1 and the remaining 35 percent are from supplier 2. Historical Data suggest the quality rating of the two supplier are shown in the table: a) Draw a tree diagram for this experiment with the probability of all outcomes b) Given the information the part is bad, What is the probability the part came from supplier 1?

Good Parts Bad Parts

Supplier 1 98 2

Supplier 1 95 5

62


Solution:

63


Example: An insurance company rents 35% of the cars for its customers from Avis and the rest from Hertz. From past records they know that 8% of Avis cars break down and 5% of Hertz cars break down. A customer calls and complains that his rental car broke down. What is the probability that his car was rented from Avis?

64


Solution:

65


Documents

Intersection of Events & Multiplication Rule · Multiplication rule Sometimes we may need to find the probability of two events happening together. The probability of the intersection