Chapter 3wellsmat.startlogic.com/sitebuildercontent/sitebuilderfiles/hstat... · In this case, the event is one outcome from the sample space. ... In a study of 420,000 cell phones

Chapter 3

Probability

Every Day, each us makes decisions

based on uncertainty.

Should you buy an extended warranty for your new DVD player?

It depends on the likelihood that it will fail during the warranty.

Should you allow 45 min to get to 1st period or is 35 min enough?

Lesson 3-1/3-2

Fundamentals

Rare Event Rule

If, under a given assumption (such as a lottery

being fair), the probability of a particular

observed event (such as five consecutive lottery

wins) is extremely small, we conclude that the

assumption is probably not correct

Probability

The probability of an event is the proportion of

times the event is expected to occur in repeated

experiments.

Sample Space (S)

The set of all possible outcomes of an event is the sample space S of the event.

Example: For the event “roll a die and observe what number it lands on” the sample space contains all possible numbers the die could land on.

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

Event

An event is an outcome (or a set of outcomes) from a sample space

Example 1: When flipping three coins, an event may be getting the result HTH. In this case, the event is one outcome from the sample space.

Example 2: When flipping three coins, an event may be getting two tails. In this case, the event is a set of outcomes (HTT, TTH, THT, …) from the sample space

An event is usually denoted by a capital letter.

Example: Call getting two tails event A.

The probability of event A is denoted P(A).

Probability Properties

P denotes probability

The probability of an event, say event A, is denoted P(A).

All probabilities are between 0 and 1.

(i.e. )

The sum of the probabilities of all possible outcomes must be 1.

The probability of an impossible event is 0.

The probability of an event that is certain to occur is 1.

0 ( ) 1P A

Assigning Probabilities to Events

Rule 1: Relative frequency approach

Rule 2: Classical approach

Rule 3: Personal opinion approach (subjective)

Rule #1 – Relative Frequency

If an experiment is repeated n times under essentially

identical conditions, and if the event A occurs m times,

then as n grows large the ratio of m/n approaches a fixed

limit, namely, the probability of A.

( )P A number of times A occurred

number of times the trial was repeated

Example – Relative Frequency

P(tack lands pointed down)

we must repeat the

procedure of tossing the tack

many times and then find the

ratio of the number of times

the tack lands with the point

down to the number of

tosses.

Rule #2 – Classical Approach

Assume that a given procedure has n different simple

events and that each those simple events has a equal

chance of occurring. If event A can occur in s of these

n ways then:

( )P A number of ways A can occur

number of different simple events

s

n

Example – Classical

When trying to determine

P(2) with a balanced and

fair dice, each of the six

faces has an equal chance

of occurring

1(2)

6P

Rule #3 – Subjective

P(A), the probability of event A, is found by simply guessing

or estimating its value based on knowledge of the relevant

circumstances.

Example – Page 128, #6

In a study of 420,000 cell phones users in Denmark, it was

found that 135 developed cancer of the brain or nervous

system. Estimate the probability that a randomly selected

cell phone user will develop such a cancer. Is the result

very different from the probability 0.000340 that was found

for the general population? What does this suggest about

cell phones as a cause of such cancers, as has been claimed?

135

( ) 0.000321420,000

P cancer

No, this is not very different from the general population value

of 0.000340. This suggests that cell phones are not the cause

of cancer.

Law of Large Numbers

As a procedure is repeated again and again, the

relative frequency (from Rule 1) of an event

tends to approach the actual probability.


Identifying Probability Values

A. What is the probability of event that is certain to occur?

P(A) = 1

B. What is the probability of an impossible event?

P(A) = 0


C.A sample space consists of 10 separate events that

are equally likely. What is the probability of each?

D. On a true/false test, what is the probability of answering

the question correctly if you make a random guess?

1( )

10P A

1( )

2P A


E. On a multiple-choice test with five possible answers for

each questions, what is the probability of answering

the questions correctly, if you make a random guess?

1( )

5P A


Probability of a Wrong Result. Table 3-1 shows that

among 14 women who were not pregnant, the test

for pregnancy yielded the wrong conclusions 3 times.

Table 3-1 Pregnancy Test Results

Subject Test Result

Pregnancy is indicated

Subject Test Result

Pregnancy not indicated

Subject is

Pregnant

80 5

Subject

Is not Pregnant

3 11


A. Based on the available results, find the probability

of wrong test conclusions for a women who is not

pregnant.

3( ) .214

14P W

Let W = The test wrongly concludes a woman is not pregnant





B. Is it “unusual” for the test conclusion to be wrong for

women who are not pregnant?

3( ) .214

14P W




No, since 0.214 > 0.05, it is not unusual for the test to

be wrong for women who are not pregnant.

Complement

The complement of an event A, denoted by Ac or Ā , is the set of outcomes that are not in A. Ac means A does not occur

P(Ac) = 1 – P(A)

Example: When flipping two coins, the probability of getting two heads is 0.25. The probability of not getting two heads is

P(Ac) = 1 – P(A) =1 – 0.25 = 0.75

Odds

Actual odds against event A occurring

Is the ratio P(Ac)/P(A), usually expressed in the form of a:b (or “a to b”), where a and b are integers having no common factors.

Actual odds in favor of a event A

Is the reciprocal of the actual odds against that event P(A)/P(Ac), then odds in favor of A are b:a.

Payoff odds against event A

Represents the ratio of net profit (if you win) to the amount bet.

Payoff odds against event A = (net profit) : (amount bet)


A roulette wheel has 38 slots. One slot is 0, anotheris 00, and the others are numbered 1 through 36respectively. You are placing a bet that the outcomeis an odd number.

A. What is the probability of winning?

18( ) 0.474

38P odd


A roulette wheel has 38 slots. One slot is 0, anotheris 00, and the others are numbered 1 through 36respectively. You are placing a bet that the outcomeis an odd number.

B. What are the actual odds against winning?

actual odds against odd =P(not odd)

P(odd)=

20

38

18

3820 38 20 10

10 : 938 18 18 9


C.When you bet the outcome is an odd number,the payoff odds are 1:1. How much profit doyou make if you bet $18 and win?

1:1 means a profit of $1 for every $1 bet.

A winning bet of $18 means a profit of $18.


D. How much profit would you make if on the $18bet if you could somehow convince the casino to change its payoff odd so they are the sameas the actual odds against winning?

Payoff odds of 10:9 mean a profit of $10 forevery $9 bet.

A winning bet of $18 means a profit of $20.

Lesson 3-3

Addition Rule

Compound Event

A compound event is any event combining two or more simple events.

Compound probability are used to answer the following question:

In rolling a single dice what is the probability of “rolling even number” or “rolling a 1 or 2”

When finding the probability that event A occurs

or event B occurs, find the total number of ways

A can occur and the number of ways B can

occur, but find the total in such a way that no

outcome is counted more than once.

Addition Rule

The addition rule is used to find probabilities involving the word “or”.

Rule – For any two events A and B

P(A P) = (A or B) = P(A) + P(B) – P(A and B)

where P(A and B) denotes the probability that Aand B both occur at the same time as an outcome in a trial or procedure.

Addition Rule – Venn Diagram


Refer to figure 3-3. Find the probability of randomly selecting one of the peas and getting one with a yellow pod or a purple flower.

P(Y or P) = P(Y) +P(P) – P(Y and P)

6 9 40.786

14 14 14


If one of the Titanic passengers is randomly selected,find the probability of getting a man or someonewho survived the sinking.

P(M or S) = P(M) + P(S) – P(M and S)

1692 706 3320.929

2223 2223 2223

Men Women Boys Girls Total

Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Mutually Exclusive

If events A and B have no simple events in common or cannot occur simultaneously, they are said to be disjoint or mutually exclusive.

For two disjoint events A and B, the probability that one or the other occurs is the sum of the probabilities of the two events.

P(A or B) = ( ) ( ) ( )P A B P A P B


Determine whether events are disjoint. Are the twoevents disjoint for a single trial?

a) Randomly selecting a head of household watching NBC on television at 8:15 tonight.

Randomly selecting a head of household watching CBS on television at 8:15 tonight.

Yes



b) Receiving a phone call from a volunteer survey subject who opposes all government taxation.

Receiving a phone call from a volunteer survey subject who approves all government taxation.

Yes



c) Randomly selecting a United States Senator currently holding office

Randomly selecting a female elected official

No

Mutually Exclusive – Venn Diagram

Applying the Addition Rule

Rule of Complementary Events

Complement Rule is used when you know the probability that some event will occur and you want to know the opposite: the chance it will not occur.

Rules

P(A) + P(AC) = 1

P(AC) = 1 – P(A)

P(A) = 1 – P(AC)

P(A) and P(AC) are mutually exclusive

All simple events are either in A or AC.

Complement – Venn Diagram


Finding the complements

A. Find given that ( )CP A ( ) 0.0175P A

( ) 1 0.0175 0.9825CP A


B. A Reuters/Zogby poll showed that 61% ofAmericans say they believe that life existssomewhere in the galaxy. What is the probability of randomly selecting someone not having that belief?

Let B = selecting an American who believesthat life exists elsewhere in the galaxy

( ) 1 0.61 0.39CP B


If someone is randomly selected, find the probabilitythat his or her birthday is not October. Ignore leapyears.

Let O = a person’s birthday falls in October

31( )

365P O

( ) 1 ( )

31 3341 0.915

365 365

CP O P O


Refer to the accompanying figure, which describesthe blood groups and Rh types of 100 people. In each case assume that 1 of the 100 is randomlyselected and find the indicated probability.

Group AB4 Rh+

1 Rh-

Group O39 Rh+

6 Rh-

Group A35 Rh+

5 Rh-

Group B8 Rh+

2 Rh-

P(group A or O or type Rh+)


Group AB4 Rh+

1 Rh-

Group O39 Rh+

6 Rh-

Group A35 Rh+

5 Rh-

Group B8 Rh+

2 Rh-

P(group A or O or type Rh+)

Rh Factor

+ – Total

Gro

up

A 35 5 40

B 8 2 10

AB 4 1 5

O 39 6 45

Total 86 14 100


P(group A or O or type Rh+) =

Rh Factor

+ – Total

Gro

up

A 35 5 40

B 8 2 10

AB 4 1 5

O 39 6 45

Total 86 14 100

P(A) + P(O) + P(Rh+) – P(A and Rh+) – P(O and Rh+) =

40 45 86 35 39 970.97

100 100 100 100 100 100

Lesson 3-4

Multiplication Rule: Basics

Independent

Events A and B are independentbecause the probability A does not change the probability of B.

Example: Roll a yellow die and a red die. Event A is the yellow die landing on an even number, and event B is the red die landing on an odd number.

Multiplication Rule

The multiplication rule is used to find probabilities involving the word “and”.


P(A B) = P(A and B) = P(A occurs in a 1st

trial and event B occurs in a 2nd trial

For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events.

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

Dependent If event A and event B are not independent, they are said

to be dependent.

The probability for the second event B should take into account the fact that the first event A has already occurred.

The probability that event B occurs if we know for certain that event A will occur is called conditional probability.

The conditional probability of B given A is:

P(B|A) which reads “the probability of event Bgiven event A has occurred.”


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

Applying the Multiplication Rule


Identify events as independent or dependent. For each pair of events, classify the two events asindependent or dependent.

a) Finding that you calculator is not working.

Finding that your refrigerator is not working.

Independent



b) Finding that your kitchen light is not working.

Finding that your refrigerator is not working.

Dependent



c) Drinking until your driving ability is impaired.

Being involved in a car crash.

Dependent


A new computer owner creates a password consisting of two characters. She randomly selects a letter of the alphabet for the first character and a digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) for the second character. What is the probability that her password is “K9”? Would this password be effective as a deterrent against someone trying to gain access to her computer?

Let A = Alphabet and N = Number

P(A and N) ( ) ( )P A P N

1 1 1

0.003826 10 260

P(K and 9)

Example - Page 147, #8A study of hunting injuries and the wearing of“hunter” orange clothing showed that among

123 hunters injured when mistaken for game, 6 werewearing orange. If a follow-up study begins withthe random selection of hunters from this sampleof 123, find the probability that the first two selected hunters were both wearing orange.

A. Assume that the first hunter is replaced beforethe next one is selected.

1 2 1

( ) ( | )P O P O O1 2

( )P O and O

6 6

0.00238123 123

Example - Page 147, #8

B. Assume that the first hunter is not replaced before the next one is selected.

1 2 1

( ) ( | )P O P O O1 2

( )P O and O

6 5

0.00200123 122

C. Given a choice between selecting with replacement and selecting without replacement,which choice makes more sense in this situation?

Selecting with replacement could lead to re-interviewing


It is common for the public opinion polls to have a “confidence level” of 95%, meaning that there is a 0.95 probability that the poll results are accurate within the claimed margins of error. If five different organizations conduct independent polls, whatis the probability that all five of them are accurate within the claimed margin of error? Does the result suggest that with a confidence level of 95%, we can expect that almost all polls will be within the claimed margin of error?

A = a poll is accurate as claimed

( ) 0.95P A for each poll

5

1 2 5( , ... ) (0.95) 0.774P A A A

No, the more polls one considers, the less likelythat they will be within the claimed margin of error.


Remote sensors are used to control each of twoseparate and independent values, denoted by p andq, that open to provide water for emergency cooling of a nuclear reactor. Each value has a 0.9968 probability of opening when triggered. For the given configuration, find the probability that when both sensors are triggered, water will get through the system so that cooling can occur. Is this result high enough to be considered safe?

Let O = the value opens properly

P(O) = 0.9968, for each value

Example – Page 148, #16Let O = the value opens properly

P(O) = 0.9968, for each value

P(water gets through) = P(O1 and O2)

= P(O1) P(O2)

= (0.9968) (0.9968)

= 0.9936


P(water gets through) = 0.9936

Is this result high enough to considered safe?

Yes; since the systems is used in emergencies, andnot on a routine basis – but because the probabilityof failure is 0.0064 (1 – 0.9936), the system can be expected to fail about 64 times in every 10,000 uses.That means if it is used 28 times a day [28 365 = 10,220 times a year], we expect about64 failures annually – and after one such failure ina nuclear reactor, there might not be a next time foranything.



Positive Negative

Yes 80 5

No 3 11

If one of the subjects is randomly selected, find theprobability of getting someone who test negative orsomeone who is not pregnant.

( ) ( )P Neg P No

16 14 11

0.19299 99 99

85

14

83 16 99

P(Neg or No) P(Neg and No)



Positive Negative

Yes 80 5

NO 3 11

If three people are randomly selected, find theprobability that they all test negative

1 2 3 1 2 1 3 1 2

( , , ) ( ) ( | ) ( | , )P N N N P N P N N P N N N

16 15 14

0.0035799 98 97

85

14

83 16 99

Lesson 3-5

Multiplication Rule

Complements and Conditional Probability

Complements: The Probability of

“At Least One”

“At least one” is equivalent to “one or more”

The complement of getting at least one item of a particular type is that you get no item of that type.

To find the probability of at least one of something, calculate the probability of none, then subtract that result from 1. That is,

P(at least one) = 1 – P(none)


Provide a written description of the complement ofthe given event.

Quality Control: When 50 HDTV units are shipped, allof them are free of defects.

At least one unit is defective.


Provide a written description of the complement ofthe given event.

A Hit with the Misses: When Mike ask five different women for a data, at least one of them accepts.

None of the women accepts.


If a couple plans to have 12 children, what is the probability that there will be at least one girl? If the couple eventually has 12 children and they are all boys, what can the couple conclude?

Let B = a child is a boyP(B) = 0.5, for each birth

P(at least one girl) = 1 – P(all boys)

1 2 121 ( ) ( ).... ( )P B P B P B

121 (.5) .999756

Either a very rare event has occurred or somegenetic factor is causing the likelihood of boy to be much greater than 0.5


If you make random guesses for four multiple-choice test questions (each with five possible answers), what is theprobability of getting at least one correct? If a very nondemanding instructor says that passing the test occursif there is a least one correct answer, can you reasonablyexpect to pass by guessing?

Let W = guessing a wrong answer4

( )5

P W

P(at least one correct) = 1 – P(all wrong)

1 2 4

1 ( ) ( )... ( )P W P W P W4

41 0.590

5


Since 0.590 > 0.50, you are more likely to passthan fail – whether or not you can “reasonablyexpect” to pass depends on your perception of

what is reasonable.

Conditional Probability

A conditional probability of event occurs when theprobability is affected by the knowledge of othercircumstances.

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

P(A)P(A)

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

P(A)


Refer to figure 3-3 in section 3-3 for the peas usedin genetics experiment. If one of the peas israndomly selected and is found to have a green pod,what is the probability that is has a purple flower.

P(Purple|Green)

=P(G and P)

P(G)

5 8 5 140.625

14 14 14 8


If we randomly select some who died, what is theis the probability of getting a man?

P(M|D)


Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

1360( and ) 13602223 0.965

1517( ) 1517

2223

P M D

P D


What is the probability of getting a man or women,given that the randomly selected person is someonewho died.?

P([M or W]|D) =P(D and [M or W])

P(D)


Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223


P([M or W]|D) =P(D and [M or W])

P(D)

[1360 104][1360 104] 14642223 0.965

1517 1517 1517

2223


Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Testing for Independence

In Section 3-4 we stated that events A and B are

independent if the occurrence of one does not

affect the probability of occurrence of the other.

This suggests the following test for independence:

Two events A and B are Independent if:

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

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

Two events A and B are dependent if:

P(B|A) ≠ P(B) or

P(A and B) ≠ P(A) P(B)

Lesson 3-6

Probabilities Through Simulations

Simulation

A simulation of a procedure is a process that behaves the same way as the procedure, so that similar results are produced.

The numbers used in a simulation must be generated in such a way that they are equally likely

Obtaining Randomly Generated

Numbers

A table of random digits

TI-83 graphing calculator


Assume that you want to use the digits in the accompanying list to simulate the rolling of singedie. If the digits 1, 2, 3, 4, 5, 6 are used while allother digits are ignored, list the outcomes from thefirst two rows.

46196994387211344044867630015164703

78907191556764098746299108285525259

147528544675260925328733355848

4, 6, 1, 6, 4, 3

Example – Page 160, #8When Mendel conducted his famous hybridizationexperiments, he used peas with green pods, andyellow pods. One experiment involved crossingpeas in such a way that 25% of the offspring peaswere expected to have yellow pods. Use the randomdigits in the margin to develop a simulation for findingthe probability that when two offspring peas areproduced, at least one of them has a yellow pods.How does the result compare to the correct probabilityof 7/16. (Hint: Because 25% of the offspring areexpected to have yellow pods and the other 75% areexpected to have green pods, let digit 1 representyellow pods and let digits 2, 3, 4 represent green podsand ignore any other digits)


46196994387211344044867630015164703

78907191556764098746299108285525259

147528544675260925328733355848

1 – yellow pods2, 3, 4 – green pods

41, 43, 21, 13, 44, 44, 31, 14, 31, 14,42, 12, 22, 14, 24, 42, 22, 32, 33, 34


1 – yellow pods2, 3, 4 – green pods

41, 43, 21, 13, 44, 44, 31, 14, 31, 14, 42, 12, 22, 14, 24, 42, 32, 33, 34

Find the number of yellow pods

0 – no yellow1 – yellow

2 – yellow

1, 0, 1, 1, 0, 0 , 1, 1, 1, 1, 01, 0, 1, 0, 0, 0, 0, 0, 0


Find the number of yellow pods 0 – no yellow1 – yellow2 – yellow

1, 0, 1, 1, 0, 0 , 1, 1, 1, 1, 01, 0, 1, 0, 0, 0, 0, 0, 0

x Frequency Relative Frequency

0 11 11/20 = 0.55

1 9 9/20 = 0.45

2 0 0/20 = 0.00

Total 20 1.00

From the simulation we estimate P(x ≥ 1) ≈ 9/20 = 0.45

This compares very favorable with 7/16 = 0.437.


Simulating Three Dice: In exercise 6 we used thedigits in the margin to simulate the rolling of dice. Simulate the rolling of three dice 100 times. Describethe simulation, then use it to estimate the probabilityof getting a total of 10 when three dice are rolled

MathPRB5:randInt( STATEDIT



P(Sum 10) = 11/100 = 0.11

This compares very favorable to the true valueof 0.125, and in this instance the simulation produced a good estimate.


In Exercise 8 we used the digits in the margin as abasis for simulating the hybridization of peas.Again assume 25% of offspring peas are expectedto have yellow pods, but develop you own simulationgenerating 100 pairs of offspring. Based on yourresults, estimate the probability of getting at leastone pea with yellow pods when two offspring are obtain.

1 – yellow pod2, 3, 4 – green pod


P(at Least 1 yellow pod) =41

0.41100

Compares favorable with the true probabilityof 0.4375

MathPRB5:randInt( STATEDIT

Lesson 3-7

Counting

Fundamental Counting Rule

For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m nways.

Example

Counting the number of possible meals

The fixed priced lunch at Sarasota High Schoolprovides the following chooses:

Appetizer: soup or saladEntrée: baked chicken, pizza, sandwich, or hotdog

Dessert: ice cream or cookie

2 4 2 16

16 different meals can be ordered.

Example – Airport Codes

The International Airline Transportation Associationuses 3 – letter codes to assign airports. How many

different airport codes

26 26 26 17,576

Example – Arranging Groups

Arrange the sequence of 6 groups.

6 5 4 3 2 1 720

Factorial Rule

A collection of n different items can be arranged

in order n! different ways.

This factorial rule reflects the fact that the first

item may be selected in n different ways, the

second item may be selected in n – 1 ways, and

so on.

Factorial symbol !

Denotes the product of decreasing positive whole numbers.

Example: 4! = 4 3 2 1 = 24

By special definition: 0! = 1

Example – Factorial

6! 6 5 4 3 2 1 720

MathPRB4:!

Number of Permutations of n Distinct

Objects Taken r at a Time

The number of arrangements of r objects chosen from n objects in which:

The n objects are distinct

once an object is used it cannot be repeated

order is important

Notation

nPr = the number of permutations of r objects selected from n objects.

Formula

n r

nP

n r

!

!

Example - Permutations

Evaluate the following

P

7 5

7! 7! 7 6 5 4 37 6 5 4 3 2,520

7 5 ! 22!

2 1

1

MathPRB2:nPr


Singing legend Frank Sinatra recorded 381 songs.From a list of his top-10 songs, you must selectthree that will be sung in a medley as a tribute atthe next MTV Music Awards ceremony. The orderof the songs is important so that they fit togetherwell. If you select three of Sinatra’s top-10

songs, how many different sequences are possible.

n

r

10

3P 10 3 720

Number of Combinations of n Distinct

Objects Taken r at a Time

The number of arrangements of n objects using r ≤ n of them in which:

The n objects are distinct

once an object is used it cannot be repeated

order is not important

Notation

nCr = Represents the number of combinations n distinct objects taken r at a time

Formula

n r

nC

r n r

!

! !

Example - Combinations

Evaluate the following

C

6 4

6! 6! 6 5 3015

4! 6 4 ! 4! 2! 2 1! 2

4!

4

MathPRB2:nCr


Find the probability of winning the New York Take Five: the winning five numbers from 1, 2,..,39

n

r

39

5C 39 5 575,757

P W 1

( )575,757

Since only one combination wins


Five Card Flush: A standard deck of cards contains13 clubs, 13 diamonds, 13 hearts, and 13 spades.If five cards are randomly selected, find the probabilityof getting a flush. (A flush is obtained when all fivecards are of the same suit. That is, they are all clubs,or all diamonds, or all hearts or all spades.

Method #1 Counting Technique

The total number of possible 5 card selections is

52 5

52!2,598,960

47!5!C


The total number of possible 5 card selections fromone particular suit is

13 5

13!1287

8!5!C

The total number of possible 5 card selections fromany of the 4 suits is

4 1287 5148

P(getting a flush) 5148

0.001982598960


Method #2 Probability Formulas

Let F = selecting a card favorable for getting a flush.

1

52( )

52P F since the 1st card could be anything

2

12( )

51P F

since the 2nd card must be from the same suit as the first

P(getting a flush) = P(F1 and F2 and F3 and F4 and F5)

= P(F1) P(F2) P(F3) P(F4) P(F5)

52 12 11 10 90.00198

52 51 50 49 48

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Chapter 3wellsmat.startlogic.com/sitebuildercontent/sitebuilderfiles/hstat... · In this case, the event is one outcome from the sample space. ... In a study of 420,000 cell phones