Section 11.8

Section 11.8The Multiplication Rules and

Conditional Probability

Math in Our World

Learning Objectives

Find the probability of two or more independent events all occurring.

Find the probability of two or more dependent events all occurring.

Find conditional probabilities.

Independent Events

Two events A and B are independent if the fact that A occurs has no effect on the probability of B occurring.

The probability of consecutive events depends on whether the first event affects the second.

An example of independent events could be:Rolling a die and getting a 6, and then rolling a second die and getting a 3.

Dependent Events

Two events A and B are dependent if the outcome of A has some effect on the probability of B occurring.

Examples of dependent events could be:Drawing a card from a deck, not replacing it, and then drawing a second card.Parking in a no-parking zone and getting a parking ticket.

Multiplication Rule 1

When two events A and B are independent, the probability of both occurring is

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

EXAMPLE 1 Using Multiplication Rule 1

A coin is flipped and a die is rolled. Find the probability of getting heads on the coin and a 4 on the die.

The two events, the coin landing on heads and the die showing 4, are independent.

SOLUTION

The probability of heads is 1/2 and the probability of getting 4 with one die is 1/6 . Using the multiplication rule:

(heads and 4) (heads) (4)P P P 1 12 6

1

12


As part of a psychology experiment on perception and memory, colored balls are picked from an urn. The urn contains three red balls, two green balls, and five white balls. A ball is selected and its color is noted. Then it is replaced. A second ball is selected and its color is noted.Find the probability of each of these.(a) Selecting two green balls.(b) Selecting a green ball and then a white ball.(c) Selecting a red ball and then a green ball.


Remember, selection is done with replacement, which makes the events independent.(a) The probability of selecting a green ball on each trial is 2/10 , so

SOLUTION

(green and green) (green) (green)P P P 2 2

10 10

4100

1

25


(b) The probability of selecting a green ball is 2/10 and the probability of selecting a white ball is 5/10 , so

SOLUTION

(green and white) (green) (white)P P P 2 5

10 10

10100

1

10

(c) The probability of selecting a red ball is 3/10 and the probability of selecting a green ball is 2/10 , so

(red and green) (red) (green)P P P 3 2

10 10

6100

3

50

EXAMPLE 3 Finding Probabilities for Three Independent Events

Three cards are drawn from a deck. After each card is drawn, its denomination and suit are noted and it is replaced before the next card is drawn. Find the probability of getting(a) Three kings.(b) Three clubs.

EXAMPLE 3 Finding Probabilities for Three Independent Events

(a) Since there are four kings, the probability of getting a king on each draw is 4/52 or 1/13 . The probability of getting three kings is

SOLUTION

1 1 113 13 13

1

2,197

(b) Since there are 13 clubs, the probability of getting a club is 13/52 or 1/4 . The probability of getting three clubs in a row is

1 1 14 4 4

164

EXAMPLE 4 Using Multiplication Rule 1 with Large Samples

According to a Gallup Poll in 2007, 47 percent of all parents in America with children under 18 feel frequent stress. If three parents are chosen randomly, find the probability that all three will say they feel frequent stress.

Let S denote the event that the chosen parent feels stress.P(S and S and S) = P(S) x P(S) x P(S)

= (0.47)(0.47)(0.47)= 0.104

SOLUTION

Multiplication Rule 2

When we are interested in finding the probability of consecutive events that are dependent, we can still use the multiplication rule, but with a minor modification.

( and ) ( ) ( given that has already occurred)P A B P A P B A

When two events A and B are dependent, the probability of both occurring is


An appliance store gets a shipment of 25 plasma TVs, and 3 of them are defective. If two of the TVs are chosen at random, find the probability that both are defective. (The first TV is not replaced after it’s tested.)


Since there are 3 defective TVs out of a total of 25, the probability of the first being defective is 3/25.After the first one is found to be defective and not replaced, there are 2 defective sets left out of 24, so the probability of the second being defective given that the first one is defective is 2/24.

SOLUTIONst nd st nd st(1 defective and 2 defective) (1 ) (2 given 1 )P P P

325

6

600

1100

2

24

EXAMPLE 6 Using Multiplication Rule 2 with Three Events

Three cards are drawn from an ordinary deck and not replaced. Find the probability of(a) Getting three jacks.(b) Getting an ace, a king, and a queen in order.(c) Getting a club, a spade, and a heart in order.(d) Getting three clubs.


SOLUTION

( ) (three jacks)a P 4 3 2

52 51 50

15,525

1 1 1

13 17 25

( ) (ace and king and queen)b P 4 4 4

52 51 50

816,575

1 4 213 51 25


SOLUTION

( ) (three clubs)d P 13 12 1152 51 50

11850

1 4 114 17 50

( ) (club and spade and heart)c P 13 13 1352 51 50

16910,200

1 13 134 51 50

Conditional Probability

( and )( )( )

P A BP B AP A

We know that to find the probability of two dependent events occurring, it’s important to find the probability of the second event occurring given that the first has already occurred. We call this the conditional probability of event B occurring given that event A has occurred, and denote it P(B | A).

EXAMPLE 7 Finding a Conditional Probability

Suppose that your professor goes stark raving mad and chooses your final grade from A, B, C, D, F, or Incomplete totally at random. Find the probability of getting an A given that you get a letter grade higher than D.

We are asked to find P(A | letter grade higher than D).Method 1 Knowing that you got a letter grade higher than D reduces the sample space to {A, B, C}, which has three outcomes. One of them is an A, so

P(A | letter grade higher than D) =

SOLUTION

13


Method 2 With the full sample space of {A, B, C, D, F, I},SOLUTION

1(A)6

P 3(letter grade higher than D)6

P

16( letter grade higher than D)36

P A 1 66 3

13


Hate crimes are defined to be crimes in which the victim is targeted because of one or more personal characteristics, such as race, religion, or sexual orientation. The table below lists the motivation for certain crimes as reported by the FBI for 2007.

(a) Find the probability that a hate crime was racially motivated given that it was a crime against persons.

Motivation Crimes against persons

Crimes against property

Crimes against society

Race 3,031 1,686 7Religion 421 1,054 2Sexual Orientation 1,039 418 3Total 4,491 3,158 12


Hate crimes are defined to be crimes in which the victim is targeted because of one or more personal characteristics, such as race, religion, or sexual orientation. The table below lists the motivation for certain crimes as reported by the FBI for 2007.

(b) Find the probability that a hate crime was against property given that it was motivated by the victim’s sexual orientation.










(a) Since we are interested only in crimes against persons, we only need to look at that column. There were 4,491 such crimes total, and 3,031 were racially motivated, so the probability is

SOLUTION






(b) This time we are given that the crime was motivated by sexual orientation, so we only need to look at that row. There were 1,460 such crimes total, and 418 were against property, so the probability is

SOLUTION

Section 11.8

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Transcript of Section 11.8