Theoretical ProbabilitySample Space: A listing of all the possible outcomes from a chance experiment being considered. All individual outcomes in a sample space are equally likely.

The sample space is {1, 2, 3, 4, 5, 6}

P(4) = 1/6; P(1) = 1/6

1Section 4.1, Page 76

Empirical Probability

P(pick orange M&M) = 137/692 = 0.198 = 19.8%

2Section 4.1, Page 75

Theoretical ProbabilitySample Space for rolling two die

P(2) = 1/36

P(5) = 4/36 = 1/9

P(7) = 6/36 = 1/6

3Section 4.1, Page 77

Theoretical ProbabilitySample Space for Gender of 3 children.

P(3 boys) = 1/8

P(same Gender) = 2/8

P(at least one girl) = 7/8

P(at most one girl) = 4/8

4Section 4.1, Page 78

Properties of Probability

5Section 4.1, Page 79

Subjective Probability

A subjective probability is a personal judgment determined by an observer with incomplete information.

TV news says chance of rain is 70%.

Students says the chance of 4.0 is statistics is 80%.

6Section 4.1, Page 78

Law of Large NumbersRoll die 6 times in each trial, record # of 1s.

7Section 4.1, page 80

As the number of trials increases the cumulative long term frequency approaches the theoretical probability: 1/6.

Results of individual trials

(1+2)/12=.25

Probability as Odds

Las Vegas Gamblers say the odds that the Seahawks win the Super Bowl are 1 to 50.

What is the probability they win?1/(1+50) = .0196

What is probability that they will not win?50/(1+50) = .9804.

8Section 4.1, Page 81

Probability vs. Statistics

Section 4.3, Page 82 9

If a chip is drawn at random from a bag containing these chips, the probability that it will be green is 20/60 =1/3.

A sample of ten 10 is drawn from the bag. There were 3 green chips. We are 95% sure that the true proportion of green chips is between .25 and .35.

Problems

Problems, Page 94 10

Problems

11Problems, Page 95

Problems

12Problems, Page 95

a. a. What percentage of the class is not watching television on school nights?

b. If a child is picked at random from the class, what is the probability the child is not watching television on school nights?

c. What percentage of the class is watching at least 4 hours of television on school nights?

d. If a child is picked at random from the class, what is the probability the child is watching at least 4 hours of television on school nights?

Problems

13Problems, Page 95

Conditional Probability of Events

The probability of surviving sinking of the Titanic was .44.The probability of surviving sinking, given a crew member, was .09.

P (survived) = .44P (survived | crew member) = .09

Section 4.2, Page 83 14

Conditional Probability Problem

Section 4.2, Page 82 15

Conditional Probability Problem

Section 4.2, Page 83 16

Problem

17Problems, Page 95

Standard Deck of Cards

52 Total Cards4 Suites – Clubs, Diamonds, Hearts, Spades3 Face Cards in each suite – Jack, Queen, King

Problems4.91 You draw a card at random from a standard deck of 52 cards. Find each of the conditional probabilities.

a.The card is a heart, given that it is red.

b.The card is a jack, given that it is a heart

c.The card is an ace, given that it is red.

d.The card is a queen, given that it is a face card.

19Problems, Section 4.2

Problems4.92 In its monthly report, the local animal shelter states that it currently has 24 dogs and 18 cats available for adoption. Eight of the dogs and 6 of the cats are male. Find each of the conditional probabilities if an animal is selected at random.

a.The pet is male, given that it is a cat.

b.The pet is a cat, given that it is a female

c.The pet is a female, given that it is a dog.

20Problems, Secion 4.2

Probability of Not A

Given two die, find the probability that the sum of a random throw is at least 3.

P(sum at least 3) = P(sum ≥ 3) = 1 – P(sum < 3) = 1 –1/36 = 35/36.

21Section 4.3, Page 84

€

P(A) =1− P(A )

Probability of A or B

Position on Budget ProposalP(voter in favor or a republican) = (136+314+14+88)/800=552/800=0.69

P(voter in favor) = 464/800 = 0.58P(voter a republican) = 224/800 = 0.28P(voter in favor and a republican) = 136/800 = 0.17.

P(in favor or a republican= P(in favor) + P(republican) – P(in favor and republican)= 0.58 +0.28 – 0.17 = 0.69

22Section 4.3, Page 84

General Addition Rule

Problems

4.93 Real estate ads suggest that 64%of homes for sale have garages, 21 % have swimming pools, and 17% have both features. What is the probability that a home for sale has a pool or a garage?

4.94 Employment data at a large company revealed that 72% of the workers are married, 44% are college graduates, and 25% are both married and college graduates. What the probability that a person is married or a college graduate?

23Problems, Section 4.3

Problems4.95 You draw one card at random from a deck of cards. What is the probability that the card is a

a.An ace or a heart?

b.A king or a red card?

24Problems, Section 4.3

Probability of A and B

Probability a voter is in favor of budget and a Republican = 136/800 = 0.17

Using Formula:P(in favor of budget) = 464/800 = 0.58P(Republican | in favor of budget) = 136/464 = 0.2931

P(in favor of budget and a republican) = P(in favor of budget) × P(Republican | in favor of budget) =0.58 × .2931 = .17

25Section 4.3, Page 85

Problems4.13 Seventy percent of kids who visit a doctor have a fever, and 30% of the kids with a fever have sore throats. What is the probability that a kid who goes to the doctor has a fever and a sore throat?

26Problems, Section 4.3

Problems

27Problems, Page 97

Mutually Exclusive Events

28Section 4.4, Page 87

Mutually Exclusive Events

A and B = ϕ, the empty set. P(A and B) = 0A and B are disjoint or mutually exclusive events.

B and C = {(5,5)}, P(B and C) = 1/36. B and C are not disjoint events

29Section 4.4, Page 89

Special Addition Rule

The general rule for addition: P(A or B) = P(A) + P(B) – P(A and B)

If A and B are disjoint or mutually exclusive: P(A and B) = 0 P(A or B) = P(A) + P(B) P(A or B or C or …E) = P(A)+P(B) +P(C) …+P(E)

A and B are disjointP(A or B) = P(A) + P(B) = 6/36 + 3/36 = ¼

A and C are disjoint, P(A or C) = P(A) + P(C) =6/36 +6/36 = ⅓

30Section 4.4, Page 90

Problems

31Problems, Page 98

Problems4.31 The Masterfoods company says that for a large bag of candies, yellow candies made up 20%, red another 20%, orange and blue and green each 10%, and the rest are brown. If you pick a candy at random from the bag, what is the probability that

a. It is yellow or brown?

b.It is red or orange?

c.It is not green?

32Problems, Section 4.4

Problems

4.32 The American Red Cross says that about 45% of the U. S. Population has Type O Blood, 40% type A, 11% type B, and the rest Type AB. If a volunteer is selected at random, what is the probability that her blood type is

a.Type O or type B?

b.Type A or type AB?

c.Not Type A?

33Problems, Section 4.4

Independent Events

34Section 4.5, Page 90

Independent EventsA = Ace on draw.B = Ace on 2nd draw.

Draw two cards replacing the first card in the deck before drawing the second card.

P(Ace on 1st draw) = 4/52P(Ace on 2nd draw | Ace on 1st draw) = 4/52P(Ace on 2nd draw | Not Ace on 1st draw) = 4/52Since P(B) = P(B | A) = P(B | Not A) drawing two cards with replacement are independent events.

Draw two cards without replacing the first card before the second draw.

P(Ace on 1st draw) = 4/52P(Ace on 2nd draw | Ace on 1st draw) = 3/51P(Ace on 2nd draw | Not Ace on 1st draw) = 4/51Since P(B) ≠ P(B | A) ≠ P(B | Not A) drawing two cards without replacement are not independent events, but dependent events.

35Section 4.5, Page 91

Special Multiplication RuleThe General Rule for multiplication: P(A and B) = P(A)*P(B|A)

If A and B are independent then: P(B|A) = P(B): P(A and B) = P(A)*P(B) P(A and B and C and… E) = P(A)*P(B)*P(C)*…*P(E)

Example 1: What is the probability of rolling three 6 in three rolls of a die.Since each roll is independent, the probability is

Example 2: For a certain bowler, the probability of getting a strike in one roll of the ball is 0.45. What is probability that this bowler will roll a perfect game, twelve strikes in a row?Since each roll in independent of the other rolls, the probability of a perfect game is

€

1

6

⎛

⎝ ⎜

⎞

⎠ ⎟3

€

(0.45)12

36Section 4.5, Page 93

Problems

37Problems, Page 98

Problems

38Problems, Page 99

Problems

39Problems, Section 4.5

4.52 The American Red Cross says that about 45% of the U. S. Population has Type O Blood, 40% type A, 11% type B, and the rest Type AB. If three volunteers are selected at random, what is the probability that

a.All three are type A?

b.None are Type AB?

c.The first one is type A, the second one is not Type AB, and the third one is Type O?

Problems

40Problems, Section 4.5

A large airplane manufacturing company has designed an new plane. Among the thousands of parts that make up the plane, there are 30 that are mission critical. Failure of a mission critical part means that the airplane will crash. The company has designed each of these parts to be 99.9% reliable.

a.What is the probability that the plane will crash on a given flight, that is, at least one of the mission critical parts fails?

b.Would you fly on this airplane?

Summary of Probability Formulas

Equally Likely Outcomes: P(A) = n(A)/n

Complement: P(A) = 1- P(not A); P(not A) =1- P(A)

General Addition Rule: P(A or B) = P(A) + P(B) – P(A and B)

If A and B are disjoint, P(A and B) = 0Then the Special Addition Rule:Then P(A or B) = P(A) + P(B)

General Multiplication Rule: P(A and B) = P(A)×P(B|A)

If A and B are independent, P(B|A) = P(B)Then the Special Multiplication Rule:

P(A and B) = P(A)×P(B)

OddsIf the odds for A are a:b, then the odds against A are

b:a. The probability of A is a/(a+b). The probability of not A is b/(b+a)

41Chapter 4