1 chapter 6 discrete probability distributions 2 goals 1.define the terms: probability distribution...

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Chapter 6

Discrete Probability Distributions

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Goals1. Define the terms:

• Probability distribution• Random variable• Continuous probability distributions (Chapter 7)• Discrete probability distributions (Chapter 6)

2. For a discrete probability distribution, calculate:• Mean• Variance & standard deviation

3. Binomial probability distribution:• Describe the characteristics• Compute probabilities using tables in appendix A• Mean of binomial distribution• Standard deviation of binomial distribution

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So Far, & The Future…Chapter 2-4

Descriptive statistics about something that has already happened: Frequency distributions Charts Measures of central tendency (measures of location) Spread of data: dispersion

Starting with Chapter 5, 6 Inferential statistics about something that would probably

happen: Probability rules for:

Addition Multiplication

Probability distributions: Gives the entire range of values that can occur based on an

experiment Describes how likely some future event is

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Define The Terms:

• Probability distribution

• Random variable

• Discrete probability distributions

• Continuous probability distributions

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Define: Probability Distribution A listing of all the outcomes of an

experiment and the probability associated with each outcome

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Random Variable Random Variable

A quantity resulting from an experiment that, by chance, can assume different values

Because the quantitative or qualitative results from an experiment are due to chance (the future is unknown), we call the variable a random variable

Examples: Number of students absent any one day The number of people in class that say “blue” is their

favorite color The number of smiles on any given day The GDP for a country for any given month The Russell 500 value at any given second

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Continuous Probability Distributions (Chapter 7)

A listing of all the outcomes of an experiment with a continuous random variable and the probability associated with each outcome

Continuous Random Variable A variable that can assume any value within a range

Depending on the measuring instrument

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Examples Of A Continuous Probability Distribution:

The distance students travel to class The time it takes an executive to drive to work The length of an afternoon nap The length of time of a particular phone callWeight, height, air pressure, distanceAlthough, money could be Discrete, it is

considered Continuous

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Discrete Probability Distributions (Chapter 6)

A listing of all the outcomes of an experiment with a discrete random variable and the probability associated with each outcome

Discrete Random Variable A variable that can assume only certain clearly

separated values Fractions and decimals are o.k. Discrete random variable is usually the result of

counting

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Examples Of A Discrete Probability Distribution:

The number of students in a classThe number of children in a familyThe number of cars entering a carwash in a

hourNumber of home mortgages approved by

Coastal Federal Bank last weekNumber of people at a boomerang tournamentScores for a dancer at the Olympics

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Main Features Of A Discrete Probability Distribution:

The probability of a particular outcome is between 0 and 1.00 0 ≤ P(x) ≤ 1

The outcomes are mutually exclusiveThe sum of the probabilities of all the mutually

exclusive outcomes = 1.00 ΣP(x) = 1

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Example Of Discrete Probability Distribution

Consider a random experiment in which a coin is tossed three times

Let random variable = x = # of headsLet Heads = H = HeadsLet Tails = T = Tails

1st 2nd 3rd1 T T T 02 T T H 13 T H T 14 T H H 25 H T T 16 H T H 27 H H T 28 H H H 3

Sample SpacePossible results for experiment

Possible Outcome

Coin Toss Number of Heads # of Heads

x0 = 1/8 = 0.1251 = 3/8 = 0.3752 = 3/8 = 0.3753 = 1/8 = 0.125Σ = 8/8 = 1

possible values of x (number of heads) are 0,1,2,3

Probability of Outcomes P(x)

The event {two heads} occurs and the random variable = 2

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Example Of Discrete Probability Distribution

Possible Values of x (Number of Heads) are 0,1,2,3

0, 1/8

1, 3/8 2, 3/8

3, 1/8

0

1/8

2/8

3/8

4/8

0 1 2 3Number of Heads

Pro

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imil

ar t

o

The outcome of zero heads occurred onceThe outcome of one head occurred three timesThe outcome of two heads occurred three times

The outcome of three heads occurred once

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For Last Experiment (Toss Coin 3 Times)

Toss 1 = trial 1 = 2 possible outcomes Toss 2 = trial 2 = 2 possible outcomes Toss 3 = trial 3 = 2 possible outcomes Sample space = 2 * 2 * 2 = 2^3 = 8

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Mean & Standard Deviation For A Discrete Probability Distribution

• Mean (measure of central location)• Mean for probability distribution = μ (mu)

• Standard Deviation (spread of data)• Standard deviation for probability distribution = σ (sigma)

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Mean For Probability Distribution = μ (mu)

• Reports the central location of the data

• Is the long-run average value of the random variable

• Is also referred to as its expected value, E(x)

• Is a weighted average• Possible values are weighted by corresponding

probabilities

)]([ xxPμ = MeanP(x) = Probability of taking on a particular value xΣ = Add

Variables & Symbols

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Standard deviation for probability distribution = σ (sigma)

• Reports the amount of spread (variation) in the distribution

2 2[( ) ( )]x P x

σ2 = Varianceσ = Standard Deviationμ = MeanP(x) = Probability of taking on a particular value xΣ = Add

Variables & Symbols

Steps:1. Subtract the mean from each value, and square the difference2. Multiply each squared difference by its probability3. Sum the resultant products4. Take the square root of the result

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For Probability Distribution Calculate: μ & σ

Tom Jones sells cars for Feeling Good Ford

Tom sells the most cars on Saturdays

The Following Discrete Probability Distribution shows the number of cars he expects to sell on a particular Saturday:

Tom expects to sell only within a certain range of cars (Not 5, 6, 40, or 1/2 car!)"

Outcomes are mutually exclusive

Number ofCars Sold

xProbability

P(x)0 0.11 0.22 0.33 0.34 0.1

Total 1

Discrete Probability Distribution for Tom's Expected Car Sales

on Saturdays

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For Probability Distribution Calculate: μ

Number ofCars Sold

xProbability

P(x)0 0.11 0.22 0.33 0.34 0.1

Total 1 Σ =

μ =

x*P(x)

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For Probability Distribution Calculate: σ

Number ofCars Sold

xProbability

P(x) (x - μ) (x - μ) 2̂0 0.11 0.22 0.33 0.34 0.1

Σ =

σ2 =σ =

(x - μ) 2̂*P(x)

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For Probability Distribution Calculate: μ

Number ofCars Sold

xProbability

P(x)0 0.11 0.22 0.33 0.34 0.1

Total 1 Σ = 2.1

μ = 2.1

0.4

0.6

00.2

0.9

x*P(x)

2.1? Over a large number of

Saturdays, Tom expects to sell a mean of 2.1 cars

We cannot sell 2.1 cars This is why we call the

value the “expected value”

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For Probability Distribution Calculate: σNumber ofCars Sold

xProbability

P(x) (x - μ) (x - μ) 2̂0 0.1 -2.1 4.411 0.2 -1.1 1.212 0.3 -0.1 0.013 0.3 0.9 0.814 0.1 1.9 3.61

Σ = 1.2900

σ2 = 1.2900σ = 1.1358

(x - μ) 2̂*P(x)0.441

0.0030.2430.361

0.242

Standard deviation = 1.136 cars? We can use this to compared to other sales people If Sue has a mean of 2.1 cars on Saturday but her standard

deviation is .93 cars, we can conclude that there is less variability in her Saturday sales that Tom’s Saturday sales

Her values are clustered more closely around the mean

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Probability Distribution Example 2

Dan Desch, owner of College Painters, studied his records for the past 20 weeks and reports the following number of houses painted per week:

# o f H o u s e s P a i n t e d per week

(x)

W e e k s

10 5

11 6

12 7

13 2

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Number of houses painted per week, (x)

Probability, P(x)

10 .25

11 .30

12 .35

13 .10

Total 1.00


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Compute the mean number of houses painted per week:

3.11

)10)(.13()35)(.12()30)(.11()25)(.10(

)]([)(

xxPxE


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Compute the standard deviation of the number of houses painted per week:

2 2

2 2 2

2

2

2

[( ) ( )]

(10 11.3) (.25) ... (13 11.3) (.10)

0.4225 0.0270 0.1715 0.2890

0.91

0.91 0.95

x P x


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Binomial Probability Distribution:

• Describe the characteristics

• Compute probabilities using tables in appendix A

• Mean of binomial distribution

• Standard deviation of binomial distribution

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The Binomial Distribution Has The Following Characteristics:

It is a discrete probability distribution Only two possible outcomes for any one

trial of a binomial experiment (success or failure) Question can be true or false Product can be acceptable or not acceptable Customer can purchase or not purchase Citizen can vote or not vote

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The random variable is the result of counting the number of successes (this will be the horizontal axis) For 20 flips of the coin, count the number of heads For 50 boxes of cereal, count the number of boxes

that weigh less than the weight printed on the box The probability of success must be the same for

each trial (this is not the probability we will be calculating) Flip a coin, .5 probability to get a head each trial Multiple choice test with four answers, ¼ probability to

get the answer correct

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Each trail must be independent of any other trial The outcome of one trial does not affect the

outcome of any other trial There is no pattern

Example: Multiple choice exam: a,a,a, b,b,b, a,a,a, ,b,b,b…

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For An Experiment To Be Binomial It Must Satisfy The Following Conditions:

1. A random variable (x) counts the # of successes in a fixed number of trials (n)

Trials make up the experiment

2. Each trail must be independent of the previous trial

Outcome of one trial does not affect the outcome of any other trial

3. An outcome on each trial of an experiment is classified into one of two mutually exclusive categories:

1. Successor

2. Failure

4. The probability of success stays the same for each trial (so does the probability of failure)

Once these conditions are met, we can build a binomial distribution

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n = Fixed # of trialsx = # of successes we wantp = "pi" = Probability of success of each trial

For Binomial Probability DistributionWe Need To Know:

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Example Of Building A Binomial Distribution There are five flights daily from Oakland to Seattle.

Suppose the probability that any flight arrives late is .20. What is the probability that none of the flights are late today?

Exactly one flight is late today? etc…

First: Are the tests for a binomial experiment met?1. Fixed number of trials?

Yes. n = 5

2. Each trial independent? Statisticians feel comfortable answering yes

3. Success or Failure? Yes. Late or not late

4. Probability of a success the same for each trial? Yes. P(S) = .2

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Example Of Building A Binomial Distribution Second: We need the values for P(0), P(1), P(2), P(3), P(4), P(5) for

the vertical axis 0, 1, 2, 3, 4, 5 will be used for the horizontal axis

n = # of trials = # of flights = 5x = # of successes we want (0 late) = 0p = Probability of success of a late plane for each trial = 0.2

P(0)

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We Will Not Use The Formula

( ) (1 )

!( ) (1 )

( )! !

# of ways we can have x successes in n trials

# of trials

# of successes we want

("pi") probability of a success

! Factorial (3! = 3*2*1)

Remember:

x n xn x

x n x

n x

P x C

nP x

n x x

C

n

x

0 0

0! = 1

Remember: X X 1n

n nn

XX

X

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But We Want To Look At An Example:

0 5 0

5

5

5!(0) .2 (1 .2)

(5 0)!0!

5!(0) 1(.8)

5!1

(0) 1(.8)

(0) 1(.32768)

(0) .32768

The probability that there will be no late arrivals is .3277

P

P

P

P

P

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We will Use Tables In Appendix For Calculating Binomial Probabilities

Page 713

0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

0 0.774 0.59 0.328 0.168 0.078 0.031 0.01 0.002 0 0 0

1 0.204 0.328 0.41 0.36 0.259 0.156 0.077 0.028 0.006 0 0

2 0.021 0.073 0.205 0.309 0.346 0.313 0.23 0.132 0.051 0.008 0.001

3 0.001 0.008 0.051 0.132 0.23 0.313 0.346 0.309 0.205 0.073 0.021

4 0 0 0.006 0.028 0.077 0.156 0.259 0.36 0.41 0.328 0.204

5 0 0 0 0.002 0.01 0.031 0.078 0.168 0.328 0.59 0.774

= p probability of success on each trial

n = # of independent trials = 5

# o

f S

uccesses

(x)

Take These Numbers and Plot Them!

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Binomial Probability Distribution for Number of Late Flights

0, 0.328

1, 0.41

2, 0.205

3, 0.051

4, 0.006 5, 00

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 1 2 3 4 5

Number of Late Flights

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The Table In The Book Will Round, So Use The Formula And Build Your Own

Tables In Excel

0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

0 0.7737809 0.59049 0.32768 0.16807 0.07776 0.03125 0.01024 0.00243 0.00032 0.00001 0.0000003

1 0.2036266 0.32805 0.4096 0.36015 0.2592 0.15625 0.0768 0.02835 0.0064 0.00045 0.0000297

2 0.0214344 0.0729 0.2048 0.3087 0.3456 0.3125 0.2304 0.1323 0.0512 0.0081 0.0011281

3 0.0011281 0.0081 0.0512 0.1323 0.2304 0.3125 0.3456 0.3087 0.2048 0.0729 0.0214344

4 0.0000297 0.00045 0.0064 0.02835 0.0768 0.15625 0.2592 0.36015 0.4096 0.32805 0.2036266

5 0.0000003 0.00001 0.00032 0.00243 0.01024 0.03125 0.07776 0.16807 0.32768 0.59049 0.7737809



# o

f S

uccesses

(x)

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Example 2 Of Building A Binomial Distribution The West Seattle Bridge is clogged with traffic during the

morning commute 10% of the time. If you plan to travel in the morning commute seven times in the next two weeks. What is the probability that you won’t get stuck in traffic? Get stuck

four times?

First: Are the tests for a binomial experiment met?1. Fixed number of trials?

Yes. n = 7

2. Each trial independent? Yes

3. Success or Failure? Yes. Clogged or not clogged

4. Probability of a success the same for each trial? Yes. P(Clogged With Traffic) = .1

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Second: Let’s find the values for P(0), P(1), P(2), P(3), P(4), P(5),

P(6), P(7) for the vertical axis 0, 1, 2, 3, 4, 5, 6, 7 will be used for the horizontal axis

Example 2 Of Building A Binomial Distribution

n = # of trials = # of trips across bridge during the commute = 7x = # of successes (clogged with traffic) = 0p = Probability of success for each trial (a clogged bridge during the commute) = 0.1

P(0)

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We will Use Tables In Appendix For Calculating Binomial Probabilities

Page 714

0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

0 0.6983 0.4783 0.2097 0.0824 0.028 0.0078 0.0016 0.0002 0 0 0

1 0.2573 0.372 0.367 0.2471 0.1306 0.0547 0.0172 0.0036 0.0004 0 0

2 0.0406 0.124 0.2753 0.3177 0.2613 0.1641 0.0774 0.025 0.0043 0.0002 0

3 0.0036 0.023 0.1147 0.2269 0.2903 0.2734 0.1935 0.0972 0.0287 0.0026 0.0002

4 0.0002 0.0026 0.0287 0.0972 0.1935 0.2734 0.2903 0.2269 0.1147 0.023 0.0036

5 0 0.0002 0.0043 0.025 0.0774 0.1641 0.2613 0.3177 0.2753 0.124 0.0406

6 0 0 0.0004 0.0036 0.0172 0.0547 0.1306 0.2471 0.367 0.372 0.2573

7 0 0 0 0.0002 0.0016 0.0078 0.028 0.0824 0.2097 0.4783 0.6983

n = # of independent trials = 7 = p probability of success on each trial

# o

f S

uccesses (x)

Take These Numbers and Plot Them!

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Binomial Probability Distribution for Number of Traffic Jams Encountered In 7 Days

0, 0.4783

1, 0.372

2, 0.124

3, 0.0234, 0.0026 5, 0.0002 6, 0 7, 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5 6 7

Number of Traffic Jams

Pro

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Graphs: If n Remains The Same, But Increases

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Graphs: If n Remains The Same, But Increases

Binomial Probability Distribution for Number of...

0, 0 1, 0.003

2, 0.016

3, 0.054

4, 0.121

5, 0.193

6, 0.226

7, 0.193

8, 0.121

9, 0.054

10, 0.016

11, 0.003 12, 00

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5 6 7 8 9 10 11 12

Number of ...

Pro

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uccess

As approaches .5,

the graph becomes

symmetrical

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Graphs: If Remains The Same, But n Increases

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Binomial Probability Distribution for Number of...

0, 0.014781

1, 0.065693

2, 0.142334

3, 0.2003234, 0.205887

5, 0.16471

6, 0.106756

7, 0.057614

8, 0.026407

9, 0.01043210, 0.003593

0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5 6 7 8 9 10

Number of ...

Pro

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of S

uccess

Graphs: If Remains The Same, But n Increases

As n gets larger, the

graph becomes symmetrical

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Mean & Standard Deviation of the Binomial Distribution

n

n = Fixed # of trialsp = Probability of success of each trial

μ = Mean of binomial distributionσ = Standard deviation of binomial distribution

For Binomial Probability Distribution

2 (1 )n

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Mean & Standard Deviation of the Binomial Distribution

The West Seattle Bridge is clogged with traffic during the morning commute 10% of the time. If you plan to travel in the morning commute seven times in the next two weeks, what is the mean and standard deviation of this distribution?

n = # of trials = # of trips across bridge during the commute = 7p = Probability of success for each trial (a clogged bridge during the commute) = 0.1

7*.1 .7

(mean of .7 times stuck in traffic)

2 7*.1(1 .1) .794

of .794 times stuck in traffic

μ = 0.7σ = 0.793725393

Over the seven days

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Cumulative Probability Distribution

0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.950 0.599 0.349 0.107 0.028 0.006 0.001 0 0 0 0 01 0.315 0.387 0.268 0.121 0.04 0.01 0.002 0 0 0 02 0.075 0.194 0.302 0.233 0.121 0.044 0.011 0.001 0 0 03 0.01 0.057 0.201 0.267 0.215 0.117 0.042 0.009 0.001 0 04 0.001 0.011 0.088 0.2 0.251 0.205 0.111 0.037 0.006 0 05 0 0.001 0.026 0.103 0.201 0.246 0.201 0.103 0.026 0.001 06 0 0 0.006 0.037 0.111 0.205 0.251 0.2 0.088 0.011 0.0017 0 0 0.001 0.009 0.042 0.117 0.215 0.267 0.201 0.057 0.018 0 0 0 0.001 0.011 0.044 0.121 0.233 0.302 0.194 0.0759 0 0 0 0 0.002 0.01 0.04 0.121 0.268 0.387 0.31510 0 0 0 0 0 0.001 0.006 0.028 0.107 0.349 0.599

P(7) = 0.215p(x<+7) = 0.833



# o

f S

uccesses (x)

Simply Add to get P(x<=7)

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Summarize Chapter 61. Define the terms:

• Probability distribution• Random variable• Discrete probability distributions• Continuous probability distributions

2. For a discrete probability distribution, calculate:• Mean• Variance & standard deviation

3. Binomial probability distribution:• Describe the characteristics• Compute probabilities using tables in appendix A• Mean binomial distribution• Standard deviation of binomial distribution

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Alternative for standard deviation for probability distribution = σ (sigma)

2 2 2( )x P x

σ2 = Varianceσ = Standard Deviationμ = MeanP(x) = Probability of taking on a particular value xΣ = Add

Variables & Symbols

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Hypergeometric Probability Distribution Requirements for a Hypergeometric Experiment?

1. An outcome on each trial of an experiment is classified into 1 of 2 mutually exclusive categories: Success or Failure

2. The random variable is the number of counted successes in a fixed number of trials

3. The trials are not independent (For each new trial, the sample space changes

4. We assume that we sample from a finite population (number of items in population is known) without replacement and n/N > 0.05. So, the probability of a success changes for each trial

N = Number Of Items In Population n = Number Of Items In Sample = Number Of Trials S = Number Of Successes In Population X = Number Of Successes In Sample

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Hypergeometric Probability Distribution If the selected items are not returned to the population and n/N <0.05,

then the Binomial Distribution can be used as a close approximation In Excel use the HYPGEOMDIST function

Requirements for a Hypergeometric Experiment?An outcome on each trial of an experiment is classified into 1 of 2 mutually exclusive categories: Success or Failure YesThe random variable is the number of counted successes in a fixed number of trials YesThe trials are not independent (For each new trial, the sample space changes YesWe assume that we sample from a finite population (number of items in population is known) without replacement and n/N > 0.05. So, the probability of a success changes for each trial Yes Test ==> 0.045 FALSE Hypergeometric Hypergeometric Number Of Items In Population N # of Employees 110 X P(X) X P(X)

Number Of Items In Sample = Number Of Trials n # of Employees Selected for meeting 5 0 0.005376 0 0.006358

Number Of Successes In Population S # of Union members out 50 70 1 0.052269 1 0.055635

Number Of Successes In Sample X # of successes in Sample 2 2 0.19495 2 0.1947213 0.348857 3 0.340762

Success s In Union Pop P(s) 4 0.29966 4 0.298166Failure f Not In Union Pop P(f) 5 0.098888 5 0.104358

1 1

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Poisson Probability Distributions

The limiting form of the binomial distribution where the probability of success is small and n is large is called the Poisson probability distribution

The binomial distribution becomes more skewed to the right (positive) as the probability of success become smaller

This distribution describes the number of times some event occurs during a specified interval (time, distance, area, or volume)

Examples of use: Distribution of errors is data entry Number of scratches in car panels

Poisson Probability Experiment1. The Random Variable is the number of times (counting) some event occurs during a

Defined Interval Random Variable:

The Random Variable can assume an infinite number of values, however the probability becomes very small after the first few occurrences (successes)

Defined Interval: The Defined Interval some kind of “Continuum” such as:

Misspelled words per page ( continuum = per page) Calls per two hour period ( continuum = per two hour period) Vehicles per day ( continuum = per day) Goals per game ( continuum = per game) Lost bags per flight ( continuum = per flight) Defaults per 30 year mortgage period ( continuum = per 30 year mortgage

period ) Interval = Continuum

2. The probability of the event is proportional to the size of the interval (the longer the interval, the larger the probability)

3. The intervals do not overlap and are independent (the number of occurrences in 1 interval does not affect the other intervals

4. When the probability of success is very small and n is large, the Poisson distribution is a limiting form of the binomial probability distribution (“Law of Improbable Events”)

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Poisson Probability

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Variable Descriptionmu mean = pi*npi Probability of successn Sample Size = # of TrialsX Random Variable = # of successes

e2.71828182845905 use the formula =EXP(1) to generate the number e

variance variance = mustandard deviation Square Root of Variance or (pi*n)^(1/2)

In Excel use the POISSON function

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Variance = Mu Always positive skew (most of the successes are in the

first few counts (0,1,2,3…) As mu becomes larger, Poisson becomes more

symmetrical We can calculate Probability with only knowledge of Mu

and X. Although using the Binomial Probability Distribution is

still technically correct, we can use the Poisson Probability Distribution to estimate the Binomial Probability Distribution when n is large and pi is small Why? Because as n gets large and pi gets small, the Poisson

and Binomial converge


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checked luggage lost per 1000 300pi = probability 0.0003n = sample size or # of Trials 1000

mu = pi*n 0.3X P(X) P(X)

0 0.7408182206817 =POISSON(B9,$B$7,0) 0.7408182206817 =B$7^B9*EXP(1)^-B$7/FACT(B9)1 0.2222454662045 0.22224546620452 0.0333368199307 0.03333681993073 0.0033336819931 0.00333368199314 0.0002500261495 0.00025002614955 0.0000150015690 0.00001500156906 0.0000007500784 0.00000075007847 0.0000000321462 0.00000003214628 0.0000000012055 0.00000000120559 0.0000000000402 0.0000000000402

10 0.0000000000012 0.000000000001211 0.0000000000000 0.000000000000012 0.0000000000000 0.0000000000000

1.0000000000000 1.0000000000000


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Using Poisson to estimate BinomialMu = pi*n = mean # of huricanes per years period 1.5n = # of Trials) = # of years of mortgage = Contiuum = 30 Continuum = 30 year periodPi prob that hurruicane hits region during one year 0.05X 0P(X>=1) = 1 - P(0) = Prob. Of at least one Huricane during years period 0.77686984 =1-B2^B5*EXP(1)^-B2/FACT(B5)

0.77686984 =1-POISSON(B5,B2,0)

BinomialFixed # of Trials? Yes n = 30Each Trial independent? YesSuccess saty same each time? Yes pi = 0.05S/F each time? Yes Hurricane or not YesX 0P(X>=1) = 1 - P(0) = Prob. Of at least one Huricane during years period 0.785361236 =1-BINOMDIST(D14,D10,D12,0)

0.78536123606 =1-COMBIN(D10,D14)*D12^D14*(1-D12)^(D10-D14)

1 chapter 6 discrete probability distributions 2 goals 1.define the terms: probability distribution...

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