chapter 5 discrete probability distributions 1 larson/farber 4th ed

Chapter 5

Discrete Probability Distributions

1Larson/Farber 4th ed

Chapter Outline

• 5.1 Probability Distributions• 5.2 Binomial Distributions• 5.3 More Discrete Probability Distributions

Larson/Farber 4th ed 2

Section 5.1

Probability Distributions


Section 5.1 Objectives

• Distinguish between discrete random variables and continuous random variables

• Construct a discrete probability distribution and its graph

• Determine if a distribution is a probability distribution

• Find the mean, variance, and standard deviation of a discrete probability distribution

• Find the expected value of a discrete probability distribution


Random Variables

Random Variable• Represents a numerical value associated with each

outcome of a probability distribution.• Denoted by x• Examples

x = Number of sales calls a salesperson makes in one day.

x = Hours spent on sales calls in one day.


Random Variables

Discrete Random Variable• Has a finite or countable number of possible

outcomes that can be listed.• Example

x = Number of sales calls a salesperson makes in one day.

x

1 530 2 4


Random Variables

Continuous Random Variable• Has an uncountable number of possible outcomes,

represented by an interval on the number line.• Example

x = Hours spent on sales calls in one day.


x

1 2430 2 …

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution:Discrete random variable (The number of stocks whose share price increases can be counted.)

x

1 3030 2 …


1. x = The number of stocks in the Dow Jones Industrial Average that have share price increaseson a given day.

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution:Continuous random variable (The amount of water can be any volume between 0 ounces and 32 ounces)

x

1 3230 2 …


2. x = The volume of water in a 32-ouncecontainer.

Discrete Probability Distributions

Discrete probability distribution• Lists each possible value the random variable can

assume, together with its probability. • Must satisfy the following conditions:


In Words In Symbols1. The probability of each value of

the discrete random variable is between 0 and 1, inclusive.

2. The sum of all the probabilities is 1.

0 P (x) 1

ΣP (x) = 1

Constructing a Discrete Probability Distribution

1. Make a frequency distribution for the possible outcomes.

2. Find the sum of the frequencies.

3. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies.

4. Check that each probability is between 0 and 1 and that the sum is 1.


Let x be a discrete random variable with possible outcomes x1, x2, … , xn.

Exercise 1: Constructing a Discrete Probability Distribution

An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were given a score from 1 to 5, where 1 was extremely passive and 5 extremely

Score, x Frequency, f

1 24

2 33

3 42

4 30

5 21

aggressive. A score of 3 indicated neither trait. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram.


Solution: Constructing a Discrete Probability Distribution

• Divide the frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable.

24(1) 0.16

150P

33(2) 0.22

150P 42

(3) 0.28150

P

30(4) 0.20

150P

21(5) 0.14

150P

x 1 2 3 4 5

P(x) 0.16 0.22 0.28 0.20 0.14

• Discrete probability distribution:



This is a valid discrete probability distribution since

1. Each probability is between 0 and 1, inclusive,0 ≤ P(x) ≤ 1.

2. The sum of the probabilities equals 1, ΣP(x) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1.

x 1 2 3 4 5

P(x) 0.16 0.22 0.28 0.20 0.14



• Histogram

1 2 3 4 50

0.05

0.1

0.15

0.2

0.25

0.3

Passive-Aggressive Traits

Score, x

Pro

babi

lity

, P(x

)

Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome.


Mean

Mean of a discrete probability distribution• μ = ΣxP(x)• Each value of x is multiplied by its corresponding

probability and the products are added


x P(x) xP(x)

1 0.16 1(0.16) = 0.16

2 0.22 2(0.22) = 0.44

3 0.28 3(0.28) = 0.84

4 0.20 4(0.20) = 0.80

5 0.14 5(0.14) = 0.70

Exercise 2: Finding the Mean

The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean.

μ = ΣxP(x) = 2.94

Solution:


Variance and Standard DeviationDefining Formulas

Variance of a discrete probability distribution

σ2 = Σ(x – μ)2P(x)

Standard deviation of a discrete probability distribution

2 2( ) ( )x P x


Exercise 3: Finding the Variance and Standard Deviation – Defining Formula

The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the variance and standard deviation. ( μ = 2.94)

x P(x)

1 0.16

2 0.22

3 0.28

4 0.20

5 0.14


𝜎 2=∑ ¿¿

Solution: Finding the Variance and Standard Deviation – Defining Formula

• Recall μ = 2.94

2 1.616 1.3 Standard Deviation:


Score, x f P(x) (x -µ)2 (x -µ)2 * P(x)1 24

0.16 3.76 0.6022 33

0.22 0.88 0.1943 42

0.28 0.00 0.0014 30

0.20 1.12 0.2255 21

0.14 4.24 0.594

TOTAL 150 1.00 10.02 1.616

Exercise 4: Variance and Standard Deviation using Computational Formula

The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the variance and standard deviation. ( μ = 2.94)

x P(x)

1 0.16

2 0.22

3 0.28

4 0.20

5 0.14


𝜎 2=∑ [𝑥2 ∙𝑃 (𝑥 ) ]−𝜇2

Finding the Variance and Standard Deviation – Computational Formula

Score, x f P(x) x * P(x) x2 * P(x)

1 24 0.16 0.16 0.162 33 0.22 0.44 0.883 42 0.28 0.84 2.524 30 0.20 0.80 3.205 21 0.14 0.70 3.50

TOTAL 150 1.00 2.94 10.26

2 1.616 1.3 Standard Deviation:

Variance:


= 10.26 – = 1.616

Exercise 5: Finding the Variance and Standard Deviation via TI-84


L1 L2Score, x f P(x)

1 24 0.162 33 0.223 42 0.284 30 0.205 21 0.14

1-Var Stats L1, L2

Expected Value

Expected value of a discrete random variable • Equal to the mean of the random variable.• E(x) = μ = Σ xP(x)• E(x) represents the “average” value of all the

outcomes• E(x) represents the value that we would expect to get

if the trials could continue forever!• Mathematical expectation plays an important role in

an area called decision theory


Example: Finding an Expected Value

At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy one ticket. What is the expected value of your gain?


Solution: Finding an Expected Value

• To find the gain for each prize, subtractthe price of the ticket from the prize: Your gain for the $500 prize is $500 – $2 = $498 Your gain for the $250 prize is $250 – $2 = $248 Your gain for the $150 prize is $150 – $2 = $148 Your gain for the $75 prize is $75 – $2 = $73

• If you do not win a prize, your gain is $0 – $2 = –$2



• Probability distribution for the possible gains (outcomes)

Gain, x $498 $248 $148 $73 –$2

P(x) 1

1500

1

1500

1

1500

1

1500

1496

1500

( ) ( )

1 1 1 1 1496$498 $248 $148 $73 ( $2)1500 1500 1500 1500 1500$1.35

E x xP x

You can expect to lose an average of $1.35 for each ticket you buy.


Exercise 6: Finding an Expected Value

• If you bet $1 in Kentucky’s Pick 4 Lottery, you either lose $1, or gain $4999.

• The winning prize is $5000, but your $1 bet is not returned, so the net gain is $4999.

• The game is played by selecting a four-digit number between 0000 and 9999.

If you bet $1 on 1234, what is your expected gain (or loss)?




To find the gain for each prize, subtract the price of the ticket from the value of the prize.

Description Payout Cost Gain (x)Fraction Decimal

Expected

Value

P(x) P(x) x*P(x)

Winner $5,000$1 $4,999 1/10000 0.0001 $0.50

Losing Ticket $0 $1 ($1)9999/10000 0.9999 ($1.00)

TOTAL 1 ($0.50)

Interpretation: In the long run, you can expect to lose an average of $ 0.50 for each $1 bet.

Section 5.1 Summary

• Distinguished between discrete random variables and continuous random variables

• Constructed a discrete probability distribution and its graph

• Determined if a distribution is a probability distribution

• Found the mean, variance, and standard deviation of a discrete probability distribution

• Found the expected value of a discrete probability distribution


Section 5.2

Binomial Distributions



• Determine if a probability experiment is a binomial experiment

• Find binomial probabilities using the binomial probability formula

• Find binomial probabilities using technology and a binomial table

• Find the mean, variance, and standard deviation of a binomial probability distribution


Binomial Experiments

1. The experiment is repeated for a fixed number of trials, where each trial is independent of other trials

2. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success (S) or as a failure (F)

3. The probability of a success P(S) is the same for each trial

4. The random variable x counts the number of successful trials


Notation for Binomial Experiments

Symbol Description

n The number of times a trial is repeated

p = P(S) The probability of success in a single trial

q = P(F) The probability of failure in a single trial q = (1 – p)

r The random variable represents a count of the number of successes in n trials: r = 0, 1, 2, 3, … , n.


Example: Binomial Experiments

Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable r.


1. A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries.

Solution: Binomial Experiments

Binomial Experiment

1. Each surgery represents a trial. There are eight surgeries, and each one is independent of the others.

2. There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F).

3. The probability of a success, P(S), is 0.85 for each surgery.

4. The random variable r counts the number of successful surgeries.



Binomial Experiment• n = 8 (number of trials)• p = 0.85 (probability of success)• q = 1 – p = 1 – 0.85 = 0.15 (probability of failure)• r = 0, 1, 2, 3, 4, 5, 6, 7, 8 (number of successful

surgeries)



Decide whether the experiment is a binomial experiment.

If it is, specify the values of n, p, and q, and list the possible values of the random variable r.


1. Flip a coin 6 times and count the number of times the coin lands on heads.

6

0.5

0,1,2,3,4,5,6


Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable r.


2. A jar contains five red marbles, nine blue marbles, and six green marbles. You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles.


Not a Binomial Experiment

• The probability of selecting a red marble on the first trial is 5/20.

• Because the marble is not replaced, the probability of success (red) for subsequent trials is no longer 5/20.

• The trials are not independent and the probability of a success is not the same for each trial.


Example: Finding Binomial Probabilities

Microfracture knee surgery has a 75% chance of success on patients with degenerative knees. The surgery is performed on three patients. Find the probability of the surgery being successful on exactly two patients.

Create a tree diagram and use it to compute this probability.


Finding Binomial Probabilities via Binary Tree

Draw a tree diagram and use the Multiplication Rule


9(2 ) 3 0.422

64P successful surgeries

Binomial Probability Formula


Binomial Probability Formula• The probability of exactly r successes in n trials is

• n = number of trials• p = probability of success• q = 1 – p probability of failure• r = number of successes in n trials

𝑃 (𝑟 )=𝑛𝐶𝑟 𝑝𝑟𝑞𝑛− 𝑟=

𝑛 !(𝑛−𝑟 )! 𝑟 !

𝑝𝑟𝑞𝑛−𝑟

Exercise 1: Finding Binomial Probabilities via Binomial Formula

Repeat the previous exercise using the binomial formula. .


2 3 2

3 2

2 1

3 1(2 )

4 4

3! 3 1

(3 2)!2! 4 4

9 1 273 0.422

16 4 64

P successful surgeries C

𝑛=3 ,𝑝=34

,𝑟=2

Exercise 2: Finding Binomial Probabilities

Use TI-84 binompdf function


TI-83/84 DISTR / 0: binompdf(n, p, r )

P(two successes ) = binompdf(3,0.75,2) =0.422

𝑛=3 ,𝑝=34

,𝑟=2

Exercise 3: Binomial Probability Distribution

• List the possible values of r with the corresponding probability of each.

• Example: Binomial probability distribution for Microfracture knee surgery: n = 3, p = Use binomial probability formula to find

probabilities.


r 0 1 2 3

P(r)

3

4

Exercise 3: Binomial Probability Distribution

Binomial Probability Distribution• Solution: Binomial probability distribution for

Microfracture knee surgery: n = 3, p = 0.75


r 0 1 2 3

P(r) 0.016 0.141 0.422 0.422

Exercise 3: Binomial Probability Distribution via TI-84

• Microfracture knee surgery: n = 3, p = 0.75


L1 L2 =binompdf(3,0.75, L1) r P(r)

0 0.0161 0.1412 0.4223 0.422

TOTAL 1.000

Example: Construct a Binomial Distribution

In a survey, workers in the U.S. were asked to name their expected sources of retirement income. Seven workers who participated in the survey are randomly selected and asked whether they expect to rely on Social


Security for retirement income. Create a binomial probability distribution for the number of workers who respond yes.

Solution: Constructing a Binomial Distribution

• 25% of working Americans expect to rely on Social Security for retirement income.

• n = 7, p = 0.25, q = 0.75, r = 0, 1, 2, 3, 4, 5, 6, 7


P(r = 0) = 7C0(0.25)0(0.75)7 = 1(0.25)0(0.75)7 ≈ 0.1335

P(r = 1) = 7C1(0.25)1(0.75)6 = 7(0.25)1(0.75)6 ≈ 0.3115

P(r = 2) = 7C2(0.25)2(0.75)5 = 21(0.25)2(0.75)5 ≈ 0.3115

P(r = 3) = 7C3(0.25)3(0.75)4 = 35(0.25)3(0.75)4 ≈ 0.1730

P(r = 4) = 7C4(0.25)4(0.75)3 = 35(0.25)4(0.75)3 ≈ 0.0577

P(r = 5) = 7C5(0.25)5(0.75)2 = 21(0.25)5(0.75)2 ≈ 0.0115

P(r = 6) = 7C6(0.25)6(0.75)1 = 7(0.25)6(0.75)1 ≈ 0.0013

P(r = 7) = 7C7(0.25)7(0.75)0 = 1(0.25)7(0.75)0 ≈ 0.0001

Solution: Constructing a Binomial Distribution


r P(r)

0 0.1335

1 0.3115

2 0.3115

3 0.1730

4 0.0577

5 0.0115

6 0.0013

7 0.0001

All of the probabilities are between 0 and 1 and the sum of the probabilities is 1.00001 ≈ 1.

Exercise 4: Finding Binomial Probabilities

A survey indicates that 41% of women in the U.S. consider reading their favorite leisure-time activity. You randomly select four U.S. women and ask them if reading is their favorite leisure-time activity. Find the probability that at least two of them respond yes.


Solution: • n = 4, p = 0.41, q = 0.59• At least two means two or more.• Find the sum of P(2), P(3), and P(4).

Solution: Finding Binomial Probabilities


P(r = 2) = 4C2(0.41)2(0.59)2 = 6(0.41)2(0.59)2 ≈ 0.351094

P(r = 3) = 4C3(0.41)3(0.59)1 = 4(0.41)3(0.59)1 ≈ 0.162654

P(r = 4) = 4C4(0.41)4(0.59)0 = 1(0.41)4(0.59)0 ≈ 0.028258

P(r ≥ 2) = P(2) + P(3) + P(4) ≈ 0.351094 + 0.162654 + 0.028258 ≈ 0.542


• Use the binomial table to find P(r )• Make histograms for binomial distributions• Compute µ and • Compute the minimum number of trials n needed to

achieve a given probability of success P(r).


Exercise 1: Finding Binomial Probabilities Using a Table

About thirty percent of working adults spend less than 15 minutes each way commuting to their jobs. You randomly select six working adults. What is the probability that exactly three of them spend less than 15 minutes each way commuting to work? Use a table to find the probability. (Source: U.S. Census Bureau)


Solution:• Binomial with n = 6, p = 0.30, r = 3

Solution: Finding Binomial Probabilities Using a Table

• A portion of Table 2 is shown


The probability that exactly three of the six workers spend less than 15 minutes each way commuting to work is 0.185.

Exercise 2: Graphing a Binomial Distribution

Fifty-nine percent of households in the U.S. subscribe to cable TV. You randomly select six households and ask each if they subscribe to cable TV. Construct a probability distribution for the random variable x. Then graph the distribution. (Source: Kagan Research, LLC)


Solution: • n = 6, p = 0.59, q = 0.41• Find the probability for each value of r

Graphing a Binomial Distribution

Solution: Graphing a Binomial Distribution


x 0 1 2 3 4 5 6

P(x) 0.005 0.041 0.148 0.283 0.306 0.176 0.042

0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Subscribing to Cable TV

Households

Pro

babi

lity

Histogram:

Mean, Variance, and Standard Deviation

• Mean: μ = np

• Variance: σ2 = npq

• Standard Deviation:


npq

Exercise 3: Finding the Mean, Variance, and Standard Deviation

In Pittsburgh, Pennsylvania, about 56% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. Interpret the results and determine any unusual values. (Source: National Climatic Data Center)


Solution: n = 30, p = 0.56, q = 0.44

Mean: μ = np = 30∙0.56 = 16.8Variance: σ2 = npq = 30∙0.56∙0.44 ≈ 7.4Standard Deviation: 30 0.56 0.44 2.7npq

Solution: Finding the Mean, Variance, and Standard Deviation


μ = 16.8 σ2 ≈ 7.4 σ ≈ 2.7

• On average, there are 16.8 cloudy days during the month of June.

• The standard deviation is about 2.7 days. • Values that are more than 2.5 standard deviations

from the mean are considered unusual: 16.8 – 2.5(2.7) =10.05, A June with 10 cloudy

days would be unusual. 16.8 + 2.5 (2.7) =23.55 A June with 24 cloudy

days would also be unusual.

Binomial Probabilities: Equivalent Formulas


Exercise 4: Compute Binomial Probabilities

If 22% of U.S. households have a Nintendo game, compute the probability that

a. exactly 5 of 12 randomly chosen families will have Nintendo games.

b. at most, 5 randomly chosen families will have Nintendo games.


Solution: Compute Binomial Probabilities


L1 L2 =binompdf(12,0.22,L1)

r P(r)

0 0.0507

1 0.1717

2 0.2663

3 0.2503

4 0.1589

5 0.0717

TOTAL 0.9696



r P(r)

5binompdf(12, 0.22, 5) = 0.0717

a) P(exactly 5 of 12 randomly chosen families will have Nintendo games)

P(5) = binompdf(12,0.22,5) = 0.0717



b) P(at most five randomly chosen families will have Nintendo games)

P(

r P(r)

0 binompdf(12, 0.22, 0) = 0.0507

1 binompdf(12, 0.22, 1) = 0.1717

2 binompdf(12, 0.22, 2) = 0.2663

3 binompdf(12, 0.22, 3) = 0.2503

4 binompdf(12, 0.22, 4) = 0.1589

5binompdf(12, 0.22, 5) = 0.0717

sum = 0.9696

Exercise 5: Use Binomial Cumulative Density Function

If 22% of U.S. households have a Nintendo game, compute the probability that at most, 5 of 12 randomly chosen families will have Nintendo games.


Solution:

P( binomcdf( 12,0.22,5) = 0.9696

Quota Problems

• We can use the binomial distribution table “backwards” to solve for a minimum number of trials.

• Given: r and p• Find: value of n that satisfies requirement for

expected value, or reliability specification• Use calculator technology or tables together with

guess/check strategy• Show minimum of three guesses and three checks

Example: Satisfy Quota for Expected Value of Distribution

Suppose a satellite requires 3 solar cells for its power, the probability that any one of these cells will fail is 0.15, and the cells operate and fail independently.

We want to find the least number of cells each satellite should have so that the expected number of working cells (the mean) is no smaller that 3. That is, n is the total number of cells, and r is the number of working cells, p is probability that a cell will work.



We want to find the least number of cells each satellite should have so that the expected number of working cells (the mean) is no smaller that 3. That is, n is the total number of cells, and r is the number of working cells, p is probability that a cell will work, and Since

Let S = “cell works” p = 0.85, q = 0.15


cell 1 cell 2 cell 3 … cell n



cell 1 cell 2 cell 3 … cell n



p= 0.85 q = 0.15 n= 4

r P(r)0 0.0011 0.0112 0.0983 0.3684 0.522

TOTAL 1.000 0 1 2 3 4 TOTAL0.000

0.100

0.200

0.300

0.400

0.500

0.600

Probability Distribution: n = 4, p = 0.85

r

P(r

)

==>



==>

Interpretive Statement: If we put up lots of satellites with four cells, we can expect that an average of 3.4 cells will work in each satellite. Note that each additional solar satellite will produce four new cells (e.g. trials) of the binomial experiment.

Exercise 6: Satisfy Reliability Quota

Suppose a satellite requires 3 solar cells for its power, the probability that any one of these cells will fail is 0.15, and the cells operate and fail independently.

We want to find the smallest number of cells the satellite should have to be 99% sure that there will be adequate power (i.e. at least three cells work).

How many cells do we need to meet the specification P(?


P(fail) = 0.15

P(fail) = 0.15

P(fail) = 0.15

. . . P(fail) = 0.15



n P(r >=3 ) = 1 - P( r <=2) Expected Successes x Y1 = 1 - binomcdf(x,0.85,2) µ = np

3 0.6141 2.55

4 0.8905 3.40

5 0.9734 4.25

6 0.9941 5.10

7 0.9988 5.95


Conclusion: We need at least 6 cells to be 99%sure that all satellites will be operational.


Interpretive Statement: In order to meet the 99% reliability specification, we will need to install at least 6 cells in each satellite to ensure that the satellite will be fully operational.

Exercise 6: Interpretive Statement


• If ___ cells are installed in a satellite, then there is a 99% probability that the satellite will be operational.

OR • 99% of all satellites with ____ cells in them will be

operational.

Exercise 6: Interpretive Statement(s)


• If 6 cells are installed in a satellite, then there is a 99% probability that the satellite will be operational.

OR • 99% of all satellites with 6 cells in them will be fully

operational.

Section 5.3 Summary

• Determined if a probability experiment is a binomial experiment

• Found binomial probabilities using the binomial probability formula

• Graphed a binomial distribution• Found the mean, variance, and standard deviation of

a binomial probability distribution• Solved quota problems


Section 5.4

More Discrete Probability Distributions



• Find probabilities using the geometric distribution• Find probabilities using the Poisson distribution• Use a Poisson Approximation to the Binomial

Distribution


Exercise 1: Geometric Distribution

Susan is taking Western Civilization on a pass/fail basis. Historically, the passing rate for this course has been 77% each term. Let n = 1, 2, 3, . . . represent the number of times a student takes this course until a passing grade is received. Assume the attempts are independent.

Construct a tree diagram for n = 1, 2, and 3 attempts. List the outcomes and their associated probabilities.



Susan is taking Western Civilization on a pass/fail basis. Historically, the passing rate for this course has been 77% each term. Use your tree diagram to compute:

a) What is the probability that Susan passes on the first try?

b) What is the probability that Susan passes on the second try?

c) What is the probability that Susan passes on the third try?

d) What is the probability that Susan passes on the nth try?




n Factors P(n)

1 0.77 0.770

2 0.23 0.77 0.177

3 0.23 0.23 0.77 0.041

Geometric Distribution

A discrete probability distribution. It satisfies the following conditions:

A trial is repeated until a success occurs

The repeated trials are independent of each other

The probability of success p is constant for each trial

• The probability that the first success will occur on trial n is P(n) = p(q)n – 1, where q = 1 – p.


Geometric Distribution


Geometric Distribution – TI-84 Calculator

The probability of success on exactly the nth try is given by


1 2 3 . . . n


The probability that a success will occur on or before the nth try is given by


1 2 3 . . .


The probability that a success will occur after try is given by =


1 2

Exercise 2 Solution: Geometric Distribution

Susan is taking Western Civilization on a pass/fail basis. Historically, the passing rate for this course has been 77% each term. Complete the following table:


Probability Description Probability Expression

TI-84 Command

success on exactly the 3rd trysuccess on or before the 3rd trysuccess after the 3rd try

Exercise 2 Solution: Geometric Distribution

Susan is taking Western Civilization on a pass/fail basis. Historically, the passing rate for this course has been 77% each term. Complete the following table:


Probability Description

Probability Expression TI-84 Command

success on exactly the 3rd try

= 0.041

success on or before the 3rd try

= 0.988

success after the 3rd try

= = 0.012


From experience, you know that the probability that you will make a sale on any given telephone call is 0.23. Find the probability that your first sale on any given day will occur on your fourth or fifth sales call.


Solution:• P(sale on fourth or fifth call) = P(4) + P(5)• Geometric with p = 0.23, q = 0.77, n = 4, 5

Solution: Geometric Distribution

• P(4) = 0.23(0.77)4–1 ≈ 0.105003• P(5) = 0.23(0.77)5–1 ≈ 0.080852


P(sale on fourth or fifth call) = P(4) + P(5)

≈ 0.105003 + 0.080852

≈ 0.186

Poisson Distribution

A discrete probability distribution. It satisfies the following conditions:

The experiment consists of counting the number of times an event occurs in a given interval

The interval can be an interval of time, area, or volume

The probability of the event occurring is the same for each interval

The number of occurrences in one interval is independent of the number of occurrences in other intervals


The Poisson Distribution

• This distribution is used to model the number of “rare” events that occur in a time interval, volume, area, length, etc…

• Examples: Number of auto accidents during a month Number of diseased trees in an acre Number of customers arriving at a bank

Poisson Distribution

• The probability of exactly r occurrences in an interval is


where e 2.71828 and λ is the mean number of occurrences in the interval

The Poisson Distribution

Exercise 4: Poisson Distribution

The mean number of accidents per month at a certain intersection is 3. What is the probability that in any given month four accidents will occur at this intersection?


= , r =4

Exercise 5: Poisson Distribution

On average, boat fishermen on Pyramid Lake catch 0.667 fish per hour. Suppose you decide to fish the lake on a boat for 7 hours.

a) What is S ?

b) What is the expected number of fish caught over the 7-hour period?

c) Construct a table for r = 0, 1, 2, . . ., 8 fish


r 0 1 2 3 4 5 6 7 8

P(r)

Exercise 5 Solution: Poisson Distribution

On average boat fishermen on Pyramid Lake catch 0.667 fish per hour. Suppose you decide to fish the lake on a boat for 7 hours.

a) What is S? S = fish is caught

b) What is the expected number of fish caught over the 7-hour period?• = • // Expected number of fish caught in a 7 hour

period



On average boat fishermen on Pyramid Lake catch 0.667 fish per hour. Suppose you decide to fish the lake on a boat for 7 hours.

c) Construct a table for r = 0,1,2,3 … 8 fish


λ 4.7

L1 r 0 1 2 3 4 5 6 7 8

L2

P(r) = poissonpdf(4.7, L1) 0.009 0.043 0.100 0.157 0.185 0.174 0.136 0.091 0.054



Probability Description

Probability Expression TI-84 Command

At least 4 fish?

More than 7 fish?

Between 2 and 5 fish inclusive

Finding Poisson ProbabilitiesUsing the Table

• We can use Page A-16 instead of the formula:

1) Find λ at the top of the table

2) Find r along the left margin of the table

Using the Poisson Table Page A17

Recall, λ = 4

r = 0

r = 4

r = 7

Poisson Approximation to the Binomial

Exercise 6: Poisson Approximation to the Binomial

In Kentucky’s Pick 4 game, you pay $1 to select a sequence of four digits, such as 2283. If you play this game once every day, find the probability of winning exactly once in 365 days.

Solution: Consider a binomial experiment:1. 2. Because there is one winning set of numbers among the 10,000 that are possible

Example: Poisson Approximation to the Binomial

1. [n = 365]2. [ np = 0.0365]

Conclusion: We may use the Poisson distribution as an approximation to the binomial distribution. Winning is a “rare” event.

= On average, we can expect to win 0.0365 times in a 365-day interval.

Example: Poisson Approximation to the Binomial

Interpretive Statement: The probability of winning exactly once in a 365-day interval is about 0.035.

Section 5.4 Summary

• Found probabilities using the geometric distribution• Found probabilities using the Poisson distribution• Used a Poisson Approximation to the Binomial

Distribution


chapter 5 discrete probability distributions 1 larson/farber 4th ed

Documents

random variable x

probability distributions5

discrete random variables

symbolsthe probability

random variablesdecide

number of sales

possible value

number of stocks