Chapter 6 Rules of Probability

6.1 Sample spaces and events 6.2 Postulates of probabilities 6.4 Additive rules 6.5 Conditional probability 6.6 Multiplication rules 6.7 Bayes theorem

6.1 Sample Spaces and Events In a probability experiment the possible outcomes

form the samples space, S

1. Pick a card and note the suit S={spade, club, heart, diamond}2. Pick a card and note the color S={red, black}3. Pick a card and note if it is an ace S={ace, not ace}

More examples

4. Roll a die and note the dots

S={1, 2, 3, 4, 5, 6}

5. Roll 2 dice and note the sum

S={2, 3, 4, … …, 11, 12}

Event

A subset of a sample space is called an event

For example, roll one die.

Even={2, 4, 6}

Odd= {1, 3, 5}

Greater than 6=empty set=

Operations

Union

A∪B=all outcomes in A or B Intersection

A∩B=all outcomes in both A and B Complement

A’=all outcomes not in A

(like S-A)

Operations of Events

If two events have no outcomes in common they are mutually exclusive (or disjoint)

i.e., A∩ B= S={1, 2, 3, 4, 5, 6} Odd ={1, 3, 5}=A Even={2, 4, 6}=B more than 2={3, 4, 5, 6}=C

‘Odd’ and ‘Even’ are mutually exclusive;‘Even’ and ‘C’ are not mutually exclusive.

Example 6.1

Roll a red die and a white die S= {(i,j): i, j=1,2,…, 6} A={(i,j): i+j=7}={sum=7} B={(3,j): j=1,..,6}={red=3}Get A∪B A∩B A’

Another example S={52 cards in deck}

Club ∪Spade = Black

Club ∩Spade=

Red Ace= {Ace of hearts, ace of diamond}

Red’= Black

Venn Diagram

A Venn Diagram shows events as potentially intersecting circles.

e.g., R=Republican F=Female

S

R F

R U F

R ∩ F

S

R F

SR F

R F

6.2 Postulates of probabilities

Roll a die S={1, 2, 3, 4, 5, 6} P(1)=P(2)=…=P(6)=1/6

For any event A:

0 P(A) 1 P(S)=1

P(A’)=1-P(A)

Example 6.2 A={1, 2}, B={ 3, 4} A and B are disjoint.(or mutually exclusive)P(A∪B)=2/6+2/6=2/3

P(A∪B)=P(A)+P(B) if A and B are mutually exclusive

5, 6

1, 2 A

3, 4B

Example 6.3

S={52 cards}

P(ace ∪ king)=P(ace or king)

=4/52+4/52=2/13

P(rain)=0.7

then P(not rain)=1-0.7=0.3

Exercise

Which of the following pairs of events are mutually exclusive? Explain.– A driver getting a ticket for speeding and a

ticket for going through a red light.– Being foreign-born and being President of the

United States.– A person wearing black shoes and green socks.– Having rain and sunshine on the 4th of July,

2005.

Exercise

If A and B are the events that Consumer Union will rate a car stereo good or poor, P(A)=0.24 and P(B)=0.35, determine the following probabilities:– P(A’)– P(AUB)– P(A ∩B)

Exercise

Let E, T, and N be the events that a car brought to a garage needs an engine overhaul, transmission repairs, or new tires. Draw a Venn diagram for the following– E U T– E U T U N– E ∩T ∩N– E ∩T ∩N’

6.3. Probability and odds-skip

6.4 Addition Rules

For mutually exclusive events

P(A1∪A2 ∪ ••• ∪Ak)=P(A1)+P(A2)+•••+P(Ak)

P(ace or king or queen)

=P(ace)+P(king)+P(queen)

=4/52+4/52+4/52=3/13

When individual outcomes are mutually exclusive,

The probability of an event A is the sum of the probabilities for all outcomes in A

For a dice,

S={1, 2, 3, 4, 5, 6}

P(even)=P(2)+P(4)+P(6)=1/6+1/6+1/6=1/2

Ax

xPAP )()(

General Addition Rule

For any two events A and B

P(A∪B)=P(A)+P(B)–P(A∩B)

A ∩ BA B

Example 6.4

Draw a card from a deck. What is the probability of getting an ace or a spade?

P(ace or spade)

=P(ace)+P(spade) – P(ace and spade)

=4/52+13/52 –1/52

=16/52

U

6.5 Conditional Probability

Roll a die. The probability of getting any number in the set of {1,2,3,4,5,6} is 1/6 !

Try to predict the outcome when I roll the die. We

predict that any number will have the same probability of showing up.

How about if you know some prior information?

If you know the number is even

Re-evaluate the probabilities:

0 probabilities for all 3 odd numbers

equal probabilities for 3 even numbers

(each even number has a probability of 1/3)

This is called the conditional probability.

Notation: P(A|B)

The conditional probability of event A, given event B (occurs)

P(1|even)=P(3|even)=P(5|even)=0

P(2|even)=P(4|even)=P(6|even)=1/3

Conditional Probability S={1,2,3,4,5,6} E=even L=less than or equal to 3 P(E|L) is the probability of “E given L”. P(even given that point total≤3) =P(E|L)=1/3.

4, 6, 1, 325

E L

# of points in E and L( | )

# of points in L

# of points in E and L / n =

# of points in L /n

( ) =

( )

P E L

P E L

P L

Conditional Probability

Verify: P(1|even)=0

P(2|even)=1/3

)(

)()|(

BP

BAPBAP

Example 6.5 Hair Eye Probability brown blue 0.3 brown brown 0.4 blond blue 0.2 blond brown 0.1--------------------------------------------P(blue eyes|blond hair)=P(blue eyes and blond hair)/P(blond hair)=0.2/(0.2+0.1)=2/3=0.67

P(blond hair|blue eyes)= P(blond hair and blue eyes)/P(blue eyes)=0.2/(0.3+0.2)=2/5=0.40

A simpler way: tabulate the probabilities

P(blue eyes|blond hair) =0.2/0.3

Hair

Eye

brown blond Subtotal

blue 0.3 0.2 0.5

brown 0.4 0.1 0.5

Subtotal 0.7 0.3 1.0

Exercise

Candidates for a committee of 3 people include 6 women and 4 men. What is the probability that the # of women chosen is 2?

Exercise Sometimes a laboratory assistant has lunch at the cafeteria

where she works, sometimes she brings her own lunch, sometimes she has lunch at a nearby restaurant, sometimes she goes home for lunch, and sometimes she skips lunch to lose weight. If the corresponding probabilities are 0.23, 0.31, 0.15, 0.24 and 0.07, find the probabilities that she will– Have lunch at the cafeteria or the nearby restaurant;– Bring her own lunch, go home for lunch, or skip lunch altogether;– Have lunch at the cafeteria or go home for lunch;– Not skip lunch to lose weight.

Exercise

The probability that it will rain in Tucson, Arizona, on a day in mid-August, that there will be a thunderstorm on that day, and that there will be rain as well as a thunderstorm are 0.27, 0.24 and 0.15. What is the probability that there will be rain and /or a thunderstorm in Tucson on such a day?

Exercise

Roll a die. What is the probability of getting a number less than 5 if we know that it is an even number?

Exercise

The probability that Henry will like a new movie is 0.70 and the probability that Jane, his girlfriend, will like it is 0.60. If the probability is 0.28 that he will like it and she will dislike it, what is the probability that he will like it given that she is not going to like it?

6.6 Multiplication Rules

Conditional probability formula

This is equivalent to

P(A∩B)=P(A|B)P(B)

Similarly

P(A∩B)=P(B|A)P(A)

)(

)()|(

BP

BAPBAP

Example 6.6

Pick up two cards from a deck without replacement

What is the probability of getting two aces?

Method 1: like in Example 5.13

# of ways to pick 2 aces(both aces)=

# of ways to pick 2 cards

4

2 4 3 6 = 0.0045

52 52 51 1326

2

P

Method 2

A=first card is an ace B=second card is an ace

P(A)=4/52

P(B|A)=3/51

( ) (1st ace and 2nd ace)

=P(1st ace)P(2nd ace|1st ace)

=P(A)P(B|A)

4 3 = 0.0045

52 51

P A B P

Toss 2 dice A= 6 first die B= 6 second die

( ) (6 first and 6 second)

=P(6 first)P(6 second| 6 first)

1 1 1 = =

6 6 36

P A B P

For more than 2 events multiply conditional probabilities in a similar way

P(A∩B∩C)=P(A)P(B|A)P(C|A∩B )

Example 6.7 Pick up 3 cards without replacement. Find

the probability of getting 3 aces.

A=1st card is an ace; B=2nd card is an ace;

C=3rd card is an ace.

P(A)=4/52

P(B|A)=3/51

P(C|A∩B)=2/50

P(A∩B∩C)=(4/52)(3/51)(2/50)

Example 6.8

Probability of getting 4 aces in 4 cards

Independent events vs Dependent events Roll a red die and white die. A=red is a 6; B= white is a 6

Then P(A)=P(B)=1/6; P(A∩B)=1/36;

P(B|A)=P(A∩B )/P(A)=1/6 =P(B)

P(white is 6|red is 6) = P(white is 6) The probability that white is a 6 is independent of whether red is a 6. “red is a 6” and “white is 6” are independent events.

Independence

If the conditional probability of B given A is equal to the unconditional probability of B

P(B|A)=P(B)then, A and B are said to be independent

This is equivalent to P(A|B)=P(A)

Or P(A∩B )=P(B)P(A|B)=P(B)*P(A)

O.W., A and B are independent

For independent events A and B

P(A∩B )=P(A)*P(B)

P(A|B)=P(A)

P(B|A)=P(B)

If any of these 3 conditions holds, then A & B are independent.

If any of these 3 conditions does not hold, then A & B are dependent.

Example 6.9

Flip two fair coins. Probability of getting two heads?

Method 1: Sample space: {HH, HT, TH, TT}

P(HH)=1/4

Method 2: A=head on 1st flip={HH, HT}

B=head on 2nd flip={HH, TH}

Show P(B|A)=P(B)

P(HH)=P(A∩B)=P(A)P(B)=(1/2)(1/2)=1/4

Example 6.10

Pick a card, set

H=heart A=ace K=king

P(H)=13/52=1/4

P(H|K)=1/4 H and K are independent

P(A)=4/52

P(A|K)=0 A and K are dependent

Independent and mutually exclusive

They are totally different concepts! Do not mix these up.

Difference?

How to get the probability?

The solutions of most probability problems use the following facts.

For equally likely outcomes #of outcomes in A P(A)=———————————— total # of possible outcomes

P(not A)=1–P(A)

More

For mutually exclusive events P(A1 or A2 or … or Ak)= P(A1)+P(A2)+… P(Ak)

For any two events

P(A or B)=P(A∪B)=P(A)+P(B)–P(A∩B)

P(A|B)=P(A∩ B)/P(B)

More

P(A and B)=P(A∩B)=P(A)P(B|A) or P(B)P(A|B)

For independent events

P(A|B)=P(A);

P(B|A)=P(B);

P(A∩B)=P(A)P(B)

For independent events A1, A2,…, Ak

P(A1∩ A2 ∩ …∩ Ak)=P(A1)…P(Ak)

More

For any events

P(A1∩ A2 ∩ …∩ Ak)

=P(A1)P(A2|A1)…P(Ak|A1∩…∩Ak-1)

Example 6.11

Toss a coin 3 times. Find

P(2 heads)=? HHT (½)(½)(½)=1/8 HTH (½)(½)(½)=1/8 THH (½)(½)(½)=1/8 P(2 heads)=P(HHT HTH THH)=P(HHT)+P(HTH)+P(THH)=3/8∪ ∪

P(at least one head)=? =1–P(no head)=1–P(TTT)=1-1/2*1/2*1/2=1-1/8=7/8

Example 6.12Pick 5 cards

P(2 aces|1 king)= P(2 aces and 1 king)/P(1 king)

#of ways to pick up 2 aces and 1 king

#of ways to pick up 1 king

4 4 44

2 1 2

4 48

1 4

probability

4

48

1

4

4 4 44

2 1 2

Example 6.13

Deal 5 cards. What is the probability that you get 5 face cards? (J,Q, or K)

Method 1:

Method 2: multiply conditional probability

12

5# of hands with all face cards52# of hands

5

48

8

49

9

50

10

51

11

52

12

Example 6.14

Deal 5 cards, what is the probability of getting 4 aces?

48(4 aces)=

52

5

P

# of hands with 4 aces

=(# of ways to pick 4 aces)*(# of ways to pick other card)

4 48= 48

4 1

Exercise

Suppose that the probability is 0.45 that a rare tropical disease is diagnosed correctly and, if diagnosed correctly, the probability is 0.60 that the patient will be cured. What is the probability that a person who has the disease will be diagnosed correctly and cured?

Exercise

The probability that Henry will like a new movie is 0.70 and the probability that Jane, his girlfriend, will like it is 0.60. If the probability is 0.28 that he will like it and she will dislike it, what is the probability that he will like it given that she is not going to like it?

Are the two events (Henry likes the movie and Jane likes it) independent?

6.7 Bayes Theorem

A test for cancer has the following properties:

C = a patient has cancer

+ = test is positive, claiming patient has cancer

P(C)=0.001

P(+|C) =0.98

P(+|C’)= 0.01

Question 1: what % of patients test positive?

Venn diagram P(C)=0.001 P(+|C) =0.98 P(+|C’)= 0.01

P(+)=P(C∩+)+P(C’∩+) =P(C)P(+|C)+P(C’)P(+|C’) =0.001*0.98+0.999*0.01 =0.01097

C ∩ +

C +

P(C’∩+)=P(C’)P(+|C’)=0.999*0.01

=0.00999

C∩+

0.001*0.98=0.00098

Tree diagram

+

C _

+

C’ _

( ) weighted average of P(+|C) and P(+|C')

=P(C)P(+|C)+P(C')P(+|C')

P

Q2: What is the probability a patient has cancer given that patient tests positive?

P(C∩+) 0.00098 ———— = ———— =0.089 P(+) 0.01097

Most of + cases are from C’ !!!

A tree diagram helps!

Rule of total probability

P(A)=P(B)P(A|B)+P(B')P(A|B')Weighted average of P(A|B) & P(A|B’)

P(B)P(A|B)

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

Generalization: B1, B2, B3 Use this formula when Given: P(A|B) P(A|B’) P(B)Need: P(A)

B

P( B)

B’

P( B’)

A

P(A|B)

P(A|B’)

A

Bayes’ Theorem

Use this when Given: P(A|B) P(A|B’) P(B)Need: P(B|A)

(A and B)( | )

P(A)

( ) ( | ) =

( ) ( | ) ( ') ( | ')

PP B A

P B P A B

P B P A B P B P A B

Example: Insurance Company figures

A=accident in a year

AP=accident prone

Given: P(A|AP)=0.10

P(A|AP’)=0.05

P(AP)=0.2

Find P(A)

Find P(AP|A)

( ) ( ) ( | ) ( ') ( | ')

= 0.20*0.10 +0.80*0.05

=0.02+0.04

=0.06

P A P AP P A AP P AP P A AP

( ) ( ) ( | ) 0.20*0.10 0.02( | )

( ) ( ) 0.06 0.06

P AP A P AP P A APP AP A

P A P A

Bayes’ Theorem

If B1, B2,…, and Bk are mutually excusive events of which one must occur, then

1 1 2 2

( ) ( | )( | )=

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

for i=1, 2,..., or k.

i ii

k k

P B P A BP B A

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

Exercise

In a cannery, assembly lines I, II, and III account for 50%, 30%, and 20% of total output. If 0.4% of the cans from assembly line I are improperly sealed, and the corresponding percentages for assembly lines II and III are 0.6% and 1.2%.

Q1: Probability of a can being improperly sealed.

Q2: Probability of a can being sealed from line I if it is found to be improperly sealed

Exercise

Imagine that the probability is 0.95 that a certain test will correctly diagnose a person with diabetes as being diabetic, and that it is 0.05 that the test will incorrectly diagnose a person without diabetes as being diabetic. Given that roughly 10% of the population is diabetic, what is the probability that a person diagnosed as being diabetic actually has diabetes?