1

Probability

2

Today’s plan

Probability

• Notations

• Laws of probability

3

Set Notation

Sample spaceAll possible outcomes of an experiment

Example: You are playing a roulette with 100 numbersSet S consists of each of the 100 possible outcomes labeled 1,2,…100: S={1,2,…,100} This is the sample space

Examples of sample space:• Toss a coin once S={H,T}• Roll a dice once S={1,2,3,4,5,6}• Toss a coin twice S={HH,HT,TH,TT}• Gender of a baby S={boy,girl}• All letters of the alphabet S={a,b,c,…,z}

4

Set Notation

Event

A subset of the sample space

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

Subset A is the odd numbers A={1,3,5}

A is a subset of S: AS

• 3 is an element of A: 3A

5

Example of an event

Event A = getting exactly one head when tossing a coin twice

S={HH,HT,TH,TT}

A={HT,TH}

6

Venn diagram

• A tool for describing relations between sets

AB

S

7

Venn diagram

Sample space: S={1,2,3,4,5,6}

A={1,3,5}

B={2,4}

A B

S

8

Venn diagram

Sample space: S={1,2,3,4,5,6}

A={1,3,5}

B={1,2,4}

A B

S

9

= Complement of A. Includes all outcomes in S that are not in A

Also denoted by Ac

= An empty set

Complement of a set

A

?S

AA

S

10

Example:

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

A={1,3,5}

{2,4,6}A

AA

S

11

Union of sets

is the union of sets A and B.– The set of outcomes that are in A or B.– The event that either A, B, or both occur.

BA

Dice example:

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

A={1,3,5} B={1,2}

}5,3,2,1{BA

AB

S

12

Example

S={1,2,3,4,5,6,7,8,9}

A={2,4,6,8}

B={1,3,5,9}

C={1,2,3}

AUC=

AUB=

AUBc=

AUCc=

{1,2,3,4,6,8}

{1,2,3,4,5,6,8,9}

{2,4,6,7,8}

{2,4,5,6,7,8,9}

13

Intersection of sets

is the intersection of sets A and B.– The set of outcomes that are in A and B.– The event that both A and B occur.

BA

}1{BA

A BBA

S Dice example:

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

A={1,3,5} B={1,2}

14

Example

S={1,2,3,4,5,6,7,8,9}

A={2,4,6,8}

B={1,3,5,9}

C={1,2,3}

A∩B=

A∩C=

A∩Bc=

Ac∩C=

{2}

{2,4,6,8}

{1,3}

15

Disjoint sets

A and B are disjoint sets if BA

AB

S

16

Example:

• S={1,2,3,4,5,6}• A={1,2,3}• B={4,5}

BA

17

Events have probability

• S={1,2,3,4,5,6}

• A={1,2,3}

What is the probability of A?

P(A) = 3/6 = 0.5

18

Probability as a Relative Frequency

(occurrence “in the long run”)Tossing a coin:

The relative frequency of occurrences of an event A, should approach the probability P(A), as the number of trials grows (when the trials are random and independent of each other).

19

Probability as ratio of sizes

• The probability of an event A:

SAP

in elements ofnumber

Ain elements ofnumber )(

Roulette example:

A={1,13}

}100,,3,2,1{ S

100

2)( AP

20

Example:• A letter is chosen at random from the word

PROFIT. What is the probability that it is a vowel?

• S – {P,R,O,F,I,T}• A – a vowel = {O,I}

333.06

2)( AP

21

Example:A stack contains eight tickets numbered 1,1,1,2,2,3,3,3. One ticket willbe drawn at random, and its number will be noted.

a. List the sample space and assign probabilities to the elementary outcomesS={1,1,1,2,2,3,3,3}p(1)=3/8 p(2)=2/8=0.25 p(3)=3/8

b. What is the probability of drawing an odd numbered ticket?A – odd numbered ticket

75.08

6

Sin elements of #

Ain elements of #)( AP

22

Probability Rules

1. Any probability is a number between 0 and 1.

For any event A,

A=head when tossing a coin once

P(A)=0.5 for a fair coin

1)(0 AP

23

Probability Rules

2. All possible outcomes together must have probability 1.

For the sample space S,

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

p(S)=1

1SP

24

Blood types example

Blood type O A B AB

probability 0.49 0.27 0.20 ?

What is the probability of type AB blood?

S={O,A,B,AB}

All probabilities must sum to 1

0.49+0.27+0.20=0.96 p(AB)=1-0.96=0.04

25

Probability Rules

3. The probability that an event does not occur is 1 minus the probability that the event does occur.

Ac is the complement of A – the event that A does not occur

p(Ac)=1-p(A)

• Other notation for complement:

)(1 APAP

26

Example:

when throwing a dice:

A={1,3} Ac={2,4,5,6}

p(Ac) = 1-p(A) = 1-2/6 = 4/6

27

Example:

when tossing a coin:

A=head Ac=tail

p(Ac) = 1-p(A) = 1-0.5 = 0.5

28

Example:Choose an acre of land in Canada at random. The probability is 0.35 that it is forest.

What is the probability that the acre chosen is not forest?

A – forestAc – not forest

p(A)=0.35P(Ac)=1-p(A)=1-0.35=0.65

29

Probability Rules

4. The probability that a empty set occurs is zero

0P

30

Example:

• S={1,2,3,4,5,6}• A={1,2,3}• B={4,5}

P(A∩B)=?

A∩B=0

31

The additive law of probability for disjoint events

)()()( BPAPBAP

AB

When A and B are disjoint

S

32

Dice example:

A={1,3} B={1,4,6}

The additive law of probability

)()()()( BAPBPAPBAP

AB

S

BA

6

4

6

1

6

3

6

2)( BAP

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

33

Example: The additive rule• A student is randomly selected from a class.• Known facts about the class:

– 35% of the students are left-handed.– 51% of the students are juniors.– There is a probability of 0.1 of observing a left-handed

junior.

• What is the probability that the selected student is either left-handed or a junior?

34

What is the sample space?

junior Left-handed

Left-handed junior

S

35

Example – cont.

• Define the events: – A is “junior”

– B is “left-handed”

• Known: p(A)=

p(B)=

p(A B)=

• P(junior or left-handed) =

)()()()( BAPBPAPBAP

junior Left-handed

Left-handed junior

S

76.01.035.051.0

0.51

0.35

0.1

36

Blood types example

Blood type O A B AB

probability 0.49 0.27 0.20 0.04

Draw a person at random from this population.

What is the probability that the person chosen has type O blood?

Answer: p(O)=0.49

What is the probability that the person chosen has either type A or type B blood?

Answer: the 2 events “having type A”, “having type B” are disjoint since no person can have both blood types.

Addition rule p(A or B)=p(A)+p(B)=0.27+0.20=0.47

37

Blood types example - cont.

Blood type O A B AB

probability 0.49 0.27 0.20 ?

3. Mary has type B blood. She can safely receive blood transfusion from people with blood types O and B. what is the probability that a randomly chosen person can donate blood to Mary?

p(OUB)=?

The events “type O” and “type B” are disjoint, and therefore

0.690.200.49p(B)p(O)B)p(O

38

Engineering example

• Fuses are handled by one of three machines: M1, M2, M3– Machine M1 yields 200 fuses per hour– Machine M2 yields 250 fuses per hour– Machine M3 yields 350 fuses per hour

• After an hour, the fuses are mixed together, and one is selected at random.

• What is the probability that the fuse was produced by machine M3?

39

Engineering example – cont.

• Let be the event “the chosen fuse is from machine ” (i=1,2,3)

iA

iM

4375.0800

350)(

3125.0800

250)(

25.0800

200)(

3

2

1

AP

AP

AP

40

Sets: Reminder

• Set S – sample space - includes all possible outcomes

• AS (subset of S)

• = complement of A

• Intersection (A and B)

• Union (A or B)

AA

S

A

BA A BBA

S

BA

41

Probability Rules

1. Any probability is a number between 0 and 1For any event A,

2. All possible outcomes together must have probability 1:

3. The probability that an event does not occur is 1 minus the probability that the event does occur.

p(Ac)=1-p(A)

4. The probability that a empty set occurs is zero

1Sp

1p(A)0

0)p(

42

Assigning probability to equally likely outcomes

Example

Find the probability of obtaining an even number in one roll of a die

Answer:

There are 6 equally likely outcomes: 1,2,3,4,5,6

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

A={an even number is observed on the die} = {2,4,6}

5.06

3

Sin outcomes of #

Ain outcomes of #)( AP

The probability of an odd number = Ac.

P(Ac)=1-p(A)=0.5

43

Die example:

A={1,3} B={1,4,6}

The additive law of probability

)()()()( BAPBPAPBAP

AB

S

BA

6

4

6

1

6

3

6

2)( BAP

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

44

Example

Let A be the event that a student gets an A in an exam, and let F be the event that a student is female. Suppose p(A)=.25 and p(F)=.40, and p(A F)=.10.

Find p(A F).

Answer:

if we simply add p(A) and p(F) we will be counting the overlap p(A F) twice.

Therefore:

p(A F) = p(A)+p(F)-p(A F)=

=.25+.40-.1=.55

45

Example - continued

Say the probability of getting an A on the exam is .25 and the probability of getting a B is .30.Now find p(A B)

Answer:The events A and B are disjoint (also called mutually exclusive), so: p(A B) =p(A) + p(B) = .25+.30=.50

This is just a special case of the addition law, where p(A B) is 0.

46

Independence and the multiplication rule

Example - penicillin

The probability that a patient is allergic to penicillin is 0.20. Suppose this drug is administered to 2 patients. What is the probability that both patients are allergic to penicillin?

Answer:A=one of the patients is allergic to penicillin

B=the other patient is allergic to penicillin

P(A)=0.2

P(B)=0.2

What event are we looking for?

?)( BAP

47

Independence and the multiplication rule

Two events are independent if knowing that one occurs does not change the probability the other occurs.

If A and B are independent:

This is the multiplication rule for independent events.

)()()( BPAPBAP

A and B

48

Back to our penicillin example:

In our example the two events are independent, since knowing that one patient is allergic doesn’t change the probability that the other patient is allergic.

04.02.02.0p(B)p(A)B)P(A

49

Example:A general can plan a campaign to fight one major battle or three small battles. He Believes that he has probability 0.6 of winning the large battle and probability of 0.8 of winning each of the small battles. Victories or defeats in the small battles are independent. The general must win either the large battle or all three small battles to win the campaign. Which strategy should he choose?

Answer:A-winning the large battle p(A)=Bi=winning a small battle i p(Bi)=p(B1)=0.8 p(B2)=0.8 p(B3)=0.8What are we looking for?p(B1 B2 B3)=?Since B1,B2,B3 are independent p(B1 B2 B3)= p(B1)×p(B2) ×(B3)=0.83=0.512The general should choose the large battle!

0.6

0.8 for each i

50

Example - Penicillin (continued)The probability that a patient is allergic to penicillin is 0.20. Suppose the drug is administered to 3 people.(a) What is the probability that all 3 patients are allergic?(b) Find the probability that at least one is not allergic

Answer:(a) Define Ai-person i is allergic

P(A1 A2 A3) = p(A1)p(A2)p(A3)=0.23=.0008

(b) p(at least one is not allergic)=

p(1 not allergic) + p(2 not allergic) + p(3 not allergic)=….

=1-p(all three are allergic)=1-0.008=0.992

Events Ai are independent

51

When events are not independent:

Example

Drawing two aces from an ordinary deck of 52 cards:

For the first card, the probability of an ace is 4/52

If the first card is an ace, the probability that the second card is an ace is 3/51

52

Conditional probabilityExample:

Here is a two-way table for a class of 100 students, classified according to whether or not they are freshmen, and whether they live on or off campus

What is the probability that a student lives on campus? 55/100=.55Suppose a student is a freshman. What is the probability that the students lives on campus? 20/25=0.80

On Off Total

F 20 5 25

Not F 35 40 75

Total 55 45 100

53

Conditional probability

We call 0.80 a conditional probability, because a condition (being a freshmen) has been specified.

In general, we write the conditional probability of event A given event B as: p(A|B).

If

O=living on campus

F=freshman

P(O)=0.55

P(O|F)=0.80

54

Conditional probability

To see how conditional probabilities relate to ordinary ones:

F)|p(O

O FFO

S

p(F)

F)p(O

FO

55

Conditional probability

In general,

p(B)

B)p(AB)|p(A

This definition, when rearranged, gives:

B)|p(B)p(AB)p(A

Written this way, we are suggesting that B occurred first.

If A occurs first, we would use the equivalent identity:

A)|p(A)p(BB)p(A

56

Example

Suppose I pick 2 students (without replacing the first before the second is picked) at random from a class of 50 where 30 are male and 20 are female. Find the probability that

1. both are male

Denote:

M1- the first is male

M2-the second is male

49

29

50

30)M|)p(Mp(M)Mp(M 12121

57

Example

2. One is male and the other is female:

There are two possibilities for this result -

- first is male, second is female

- first is female, second is male

)Mp(F )Fp(M 2121 or

)F|)p(Mp(F)M|)p(Fp(M 121121

49

30

50

20

49

20

50

30

58

Example

Tossing a balanced coin twice: A=first toss is a head B=second toss is a head

What is the probability of obtaining head on the first and the second tosses?

A1=head on the first toss

A2=head on the second toss

p(A1 A2)=p(A1)p(A2|A1)=

=0.5×0.5=p(A1) p(A2)

When events are independent:

p(A B)=p(A) ×p(B)