1. Discrete Probability Distributions
Chapter 5
Discrete Probability Distributions
Larson/Farber 4th ed

2. Chapter Outline 5.1 Probability Distributions
5.2 Binomial Distributions
5.3 More Discrete Probability Distributions
Larson/Farber 4th ed

3. Probability Distributions
Section 5.1
Probability Distributions
Larson/Farber 4th ed

4. 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
Larson/Farber 4th ed

5. 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.
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6. 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 5 3 2 4
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7. 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 24 3 2 …
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8. Example: Random Variables
Decide whether the random variable x is discrete or continuous.
x = The number of stocks in the Dow Jones Industrial Average that have share price increases on a given day.
Solution: Discrete random variable (The number of stocks whose share price increases can be counted.) x 1 30 3 2 …
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9. Example: Random Variables
Decide whether the random variable x is discrete or continuous.
x = The volume of water in a 32-ounce container.
Solution: Continuous random variable (The amount of water can be any volume between 0 ounces and 32 ounces) x 1 32 3 2 …
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10. Discrete Probability Distributions
Lists each possible value the random variable can assume, together with its probability.
Must satisfy the following conditions:
In Words In Symbols
The probability of each value of the discrete random variable is between 0 and 1, inclusive.
The sum of all the probabilities is 1.
0  P (x)  1
ΣP (x) = 1
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11. Constructing a Discrete Probability Distribution
Let x be a discrete random variable with possible outcomes x1, x2, … , xn.
Make a frequency distribution for the possible outcomes.
Find the sum of the frequencies.
Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies.
Check that each probability is between 0 and 1 and that the sum is 1.
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12. 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
aggressive. A score of 3 indicated neither trait. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram.
Score, x
Frequency, f 1 24 2 33 3 42 4 30 5 21
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13. 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.
Discrete probability distribution: x 1 2 3 4 5
P(x)
0.16
0.22
0.28
0.20
0.14
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14. Solution: Constructing a Discrete Probability Distributionx 1 2 3 4 5
P(x)
0.16
0.22
0.28
0.20
0.14
This is a valid discrete probability distribution since
Each probability is between 0 and 1, inclusive, 0 ≤ P(x) ≤ 1.
The sum of the probabilities equals 1, ΣP(x) = = 1.
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15. Solution: Constructing a Discrete Probability Distribution
Histogram
Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome.
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16. 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
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17. Exercise 2: Finding the Mean
The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean.
Solution: 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
μ = ΣxP(x) = 2.94
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18. Variance and Standard Deviation Defining Formulas
Variance of a discrete probability distribution
σ2 = Σ(x – μ)2P(x)
Standard deviation of a discrete probability distribution
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19. 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 = 𝑥− 𝜇 2 ∙𝑃(𝑥
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20. Solution: Finding the Variance and Standard Deviation – Defining Formula
Recall μ = 2.94
Score, x f
P(x)
(x -µ)2
(x -µ)2 * P(x) 1 24
0.16
3.76
0.602 2 33
0.22
0.88
0.194 3 42
0.28
0.00
0.001 4 30
0.20
1.12
0.225 5 21
0.14
4.24
0.594
TOTAL
150
1.00
10.02
1.616
Standard Deviation:
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21. 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 = 𝑥 2 ∙𝑃 𝑥 − 𝜇 2
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22. Finding the Variance and Standard Deviation – Computational Formula
Score, x f
P(x)
x * P(x)
x2 * P(x) 1 24
0.16 2 33
0.22
0.44
0.88 3 42
0.28
0.84
2.52 4 30
0.20
0.80
3.20 5 21
0.14
0.70
3.50
TOTAL
150
1.00
2.94
10.26
Variance:
𝜎 2 = 𝑥 2 ∙𝑃 𝑥 − 𝜇 2 = – 𝟐.𝟗𝟒 𝟐 = 1.616
Standard Deviation:
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23. Exercise 5: Finding the Variance and Standard Deviation via TI-84L1 L2
Score, x f
P(x) 1 24
0.16 2 33
0.22 3 42
0.28 4 30
0.20 5 21
0.14
1-Var Stats L1, L2
𝑥 =2.9 𝜎=
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24. 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
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25. 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?
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26. Solution: Finding an Expected Value
To find the gain for each prize, subtract the 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
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27. Solution: Finding an Expected Value
Probability distribution for the possible gains (outcomes)
Gain, x
$498
$248
$148
$73
–$2
P(x)
You can expect to lose an average of $1.35 for each ticket you buy.
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28. 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)?
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29. Solution: Finding an Expected Value
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)
x*P(x)
Winner
$5,000 $1
$4,999
1/10000
0.0001
$0.50
Losing Ticket $0
($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.
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30. 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
Larson/Farber 4th ed

31. Binomial Distributions
Section 5.2
Binomial Distributions
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32. Section 5.2 Objectives
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
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33. Binomial Experiments
The experiment is repeated for a fixed number of trials, where each trial is independent of other trials
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)
The probability of a success P(S) is the same for each trial
The random variable x counts the number of successful trials
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34. 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.
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35. 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.
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.
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36. Solution: Binomial Experiments
Each surgery represents a trial. There are eight surgeries, and each one is independent of the others.
There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F).
The probability of a success, P(S), is 0.85 for each surgery.
The random variable r counts the number of successful surgeries.
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37. Solution: Binomial Experiments
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)
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38. 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. Flip a coin 6 times and count the number of times the coin lands on heads. 𝑛= 6 𝑝=
0.5 𝑟=
0,1,2,3,4,5,6
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39. 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.
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.
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40. Solution: Binomial Experiments
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.
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41. 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.
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42. Finding Binomial Probabilities via Binary Tree
Draw a tree diagram and use the Multiplication Rule
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43. Binomial Probability Formula
The probability of exactly r successes in n trials is
𝑃 𝑟 = 𝑛 𝐶 𝑟 𝑝 𝑟 𝑞 𝑛−𝑟 = 𝑛! 𝑛−𝑟 !𝑟! 𝑝 𝑟 𝑞 𝑛−𝑟
n = number of trials
p = probability of success
q = 1 – p probability of failure
r = number of successes in n trials
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44. Exercise 1: Finding Binomial Probabilities via Binomial Formula
Repeat the previous exercise using the binomial formula. .
𝑛=3, 𝑝= 3 4 , 𝑟=2
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45. Exercise 2: Finding Binomial Probabilities
Use TI-84 binompdf function
𝑛=3, 𝑝= 3 4 , 𝑟=2
TI-83/84
DISTR / 0: binompdf(n, p, r )
P(two successes ) = binompdf(3,0.75,2) =0.422
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46. 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 1 2 3
P(r)
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47. Exercise 3: Binomial Probability Distribution
Solution: Binomial probability distribution for Microfracture knee surgery: n = 3, p = 0.75 r 1 2 3
P(r)
0.016
0.141
0.422
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48. 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)
0.016 1
0.141 2
0.422 3
TOTAL
1.000
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49. 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.
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50. 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 ≈
P(r = 1) = 7C1(0.25)1(0.75)6 = 7(0.25)1(0.75)6 ≈
P(r = 2) = 7C2(0.25)2(0.75)5 = 21(0.25)2(0.75)5 ≈
P(r = 3) = 7C3(0.25)3(0.75)4 = 35(0.25)3(0.75)4 ≈
P(r = 4) = 7C4(0.25)4(0.75)3 = 35(0.25)4(0.75)3 ≈
P(r = 5) = 7C5(0.25)5(0.75)2 = 21(0.25)5(0.75)2 ≈
P(r = 6) = 7C6(0.25)6(0.75)1 = 7(0.25)6(0.75)1 ≈
P(r = 7) = 7C7(0.25)7(0.75)0 = 1(0.25)7(0.75)0 ≈
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51. Solution: Constructing a Binomial Distribution
P(r)
0.1335 1
0.3115 2 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.
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52. 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).
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53. Solution: Finding Binomial Probabilities
P(r = 2) = 4C2(0.41)2(0.59)2 = 6(0.41)2(0.59)2 ≈
P(r = 3) = 4C3(0.41)3(0.59)1 = 4(0.41)3(0.59)1 ≈
P(r = 4) = 4C4(0.41)4(0.59)0 = 1(0.41)4(0.59)0 ≈
P(r ≥ 2) = P(2) + P(3) + P(4) ≈
≈ 0.542
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54. Section 5.3 Objectives Use the binomial table to find P(r )
Make histograms for binomial distributions
Compute µ and 𝜎 for binomial distributions
Compute the minimum number of trials n needed to achieve a given probability of success P(r).
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55. 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
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56. 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
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57. 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
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58. Graphing a Binomial Distribution

59. Solution: Graphing a Binomial Distributionx 1 2 3 4 5 6
P(x)
0.005
0.041
0.148
0.283
0.306
0.176
0.042
Histogram:
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60. Mean, Variance, and Standard Deviation
Mean: μ = np
Variance: σ2 = npq
Standard Deviation:
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61. 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 = 16.8
Variance: σ2 = npq = 30∙0.56∙0.44 ≈ 7.4
Standard Deviation:
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62. Solution: Finding the Mean, Variance, and Standard Deviation
μ = σ2 ≈ 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.
(2.7) =23.55 A June with 24 cloudy days would also be unusual.
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63. Binomial Probabilities: Equivalent Formulas
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64. Exercise 4: Compute Binomial Probabilities
If 22% of U.S. households have a Nintendo game, compute the probability that
exactly 5 of 12 randomly chosen families will have Nintendo games.
at most, 5 randomly chosen families will have Nintendo games.
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65. Solution: Compute Binomial Probabilities
L2 =binompdf(12,0.22,L1) r
P(r)
0.0507 1
0.1717 2
0.2663 3
0.2503 4
0.1589 5
0.0717
TOTAL
0.9696
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66. Solution: Compute Binomial Probabilities5
binompdf(12, 0.22, 5) =
a) P(exactly 5 of 12 randomly chosen families will have Nintendo games)
P(5) = binompdf(12,0.22,5) =
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67. Solution: Compute Binomial Probabilities
binompdf(12, 0.22, 0) = 1
binompdf(12, 0.22, 1) = 2
binompdf(12, 0.22, 2) = 3
binompdf(12, 0.22, 3) = 4
binompdf(12, 0.22, 4) = 5
binompdf(12, 0.22, 5) =
sum =
b) P(at most five randomly chosen families will have Nintendo games)
P(𝑟 ≤5)=𝑃(0)+𝑃(1)+𝑃(2)+𝑃(3)+𝑃(4)+𝑃(5)=0.9696
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
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68. 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(𝑟 ≤5)=
𝑃(0)+𝑃(1)+𝑃(2)+𝑃(3)+𝑃(4)+𝑃(5)= binomcdf( 12,0.22,5) =
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
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69. 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

70. 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.
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
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71. Example: Satisfy Quota for Expected Value of Distribution
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 𝜇≥3.
Since 𝜇=𝑛𝑝, 𝑛𝑝 ≥3 𝑎𝑛𝑑 3 ≤𝑛𝑝
Let S = “cell works” p = 0.85, q = 0.15
cell 1
cell 2
cell 3 …
cell n
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
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72. Example: Satisfy Quota for Expected Value of Distribution
cell 1
cell 2
cell 3 …
cell n
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
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73. Example: Satisfy Quota for Expected Value of Distribution
0.85
q =
0.15 n= 4 r
P(r)
0.001 1
0.011 2
0.098 3
0.368
0.522
TOTAL
1.000
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
𝑛 ≥3.53 ==> choose 𝑛=4
𝜇=𝑛𝑝= =3.4
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74. Example: Satisfy Quota for Expected Value of Distribution
𝑛 ≥3.53 ==> choose 𝑛=4
𝜇=𝑛𝑝= =3.4
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.
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
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75. 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(𝑟 ≥3) ≥99%?
NOTE: If r continued up to and including 12, the probability sum would equal to 1.
P(fail) = 0.15
. . .
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76. Exercise 6: Satisfy Reliability Quotan
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
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77. Exercise 6: Satisfy Reliability Quota
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.
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78. 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.
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79. 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.
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80. 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
Larson/Farber 4th ed

81. More Discrete Probability Distributions
Section 5.4
More Discrete Probability Distributions
Larson/Farber 4th ed

82. Section 5.4 Objectives
Find probabilities using the geometric distribution
Find probabilities using the Poisson distribution
Use a Poisson Approximation to the Binomial Distribution
Larson/Farber 4th ed

83. 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.
Larson/Farber 4th ed

84. 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. Use your tree diagram to compute:
What is the probability that Susan passes on the first try?
What is the probability that Susan passes on the second try?
What is the probability that Susan passes on the third try?
What is the probability that Susan passes on the nth try?
Larson/Farber 4th ed

85. Exercise 1: Geometric Distribution
Factors
P(n) 1
0.77
0.770 2
0.23
0.177 3
0.041
𝑃 𝑛 = 𝑝 𝑞 𝑛−1
Larson/Farber 4th ed

86. 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.
Larson/Farber 4th ed

87. Geometric Distribution
Larson/Farber 4th ed

88. Geometric Distribution – TI-84 Calculator
The probability of success on exactly the nth try is given by 𝑃 𝑛 =𝑝 𝑞 𝑛−1 =𝑔𝑒𝑜𝑚𝑒𝑡𝑝𝑑𝑓(𝑝,𝑛) 1 2 3
. . . n
Larson/Farber 4th ed

89. Geometric Distribution – TI-84 Calculator
The probability that a success will occur on or before the nth try is given by
𝑃 𝑛 ≤ 𝑛 𝑚𝑎𝑥 =𝑃 1 +𝑃 2 +𝑃 𝑃 (𝑛 max⁡ )=𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(𝑝, 𝑛 max⁡ ) 1 2 3
. . .
𝑛 max⁡
Larson/Farber 4th ed

90. Geometric Distribution – TI-84 Calculator
The probability that a success will occur after try 𝑛 𝑚𝑖𝑛 is given by 𝑃 𝑛> 𝑛 𝑚𝑖𝑛 =
1 −𝑃 𝑛 ≤ 𝑛 𝑚𝑖𝑛 =1 −𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(𝑝, 𝑛 m𝑖𝑛 ) 1 2
𝑛 m𝑖𝑛⁡
> 𝑛 m𝑖𝑛⁡
Larson/Farber 4th ed

91. 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
success on or before the 3rd try
success after the 3rd try
Larson/Farber 4th ed

92. 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
𝑃 3 =(0.77 )(0.23) 2
𝑔𝑒𝑜𝑚𝑒𝑡𝑝𝑑𝑓(0.77,3) = 0.041
success on or before the 3rd try
𝑃 𝑛 ≤ 3
𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(0.77,3) = 0.988
success after the 3rd try
𝑃 𝑛> 3 =
1 −𝑃 𝑛 ≤3
1 −𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(0.77,3) = 0.012
Larson/Farber 4th ed

93. Exercise 3: Geometric Distribution
From experience, you know that the probability that you will make a sale on any given telephone call is 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
Larson/Farber 4th ed

94. Solution: Geometric Distribution
P(4) = 0.23(0.77)4–1 ≈
P(5) = 0.23(0.77)5–1 ≈
P(sale on fourth or fifth call) = P(4) + P(5) ≈
≈ 0.186
Larson/Farber 4th ed

95. 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
Larson/Farber 4th ed

96. 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

97. 𝑃 𝑟 = 𝑒 −𝜆 ∙ 𝜆 𝑟 𝑟! =𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓(𝜆,𝑟)
Poisson Distribution
The probability of exactly r occurrences in an interval is
𝑃 𝑟 = 𝑒 −𝜆 ∙ 𝜆 𝑟 𝑟! =𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓(𝜆,𝑟)
where e  and λ is the mean number of occurrences in the interval
Larson/Farber 4th ed

98. The Poisson Distribution

99. 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?
𝑢𝑛𝑖𝑡 𝑟𝑎𝑡𝑒= 3 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡𝑠 𝑚𝑜𝑛𝑡ℎ = 𝜆 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡𝑠 𝑚𝑜𝑛𝑡ℎ
 𝜆=3, r =4
 𝑃 4 = 𝑒 −3 ∙ ! =𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓 3,4 ≈0.168
Larson/Farber 4th ed

100. Exercise 5: Poisson Distribution
On average, boat fishermen on Pyramid Lake catch fish per hour. Suppose you decide to fish the lake on a boat for 7 hours.
What is S ?
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 1 2 3 4 5 6 7 8
P(r)
Larson/Farber 4th ed

101. Exercise 5 Solution: Poisson Distribution
On average boat fishermen on Pyramid Lake catch 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?
𝑢𝑛𝑖𝑡 𝑟𝑎𝑡𝑒= 𝑓𝑖𝑠ℎ ℎ𝑜𝑢𝑟 = 𝜆 𝑓𝑖𝑠ℎ 7 ℎ𝑜𝑢𝑟𝑠
𝜆= ≈ // Expected number of fish caught in a 7 hour period
Larson/Farber 4th ed

102. Exercise 5 Solution: Poisson Distribution
On average boat fishermen on Pyramid Lake catch 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 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
Larson/Farber 4th ed

103. Exercise 5 Solution: Poisson Distribution
Probability Description
Probability Expression
TI-84 Command
At least 4 fish?
𝑃 𝑟≥4 =1−𝑃(𝑟≤3)
1−𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,3 =0.690
More than 7 fish?
𝑃 𝑟>7 =𝑃 𝑟≥8 = 1−𝑃(𝑟≤7)
1−𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,7 =0.104
Between 2 and 5 fish inclusive
𝑃 2≤𝑟≤5 = 𝑃 𝑟≤5 −𝑃 𝑟≤1
𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,5 −𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,1 =0.617
Larson/Farber 4th ed

104. Finding Poisson Probabilities Using the Table
We can use Page A-16 instead of the formula:
Find λ at the top of the table
Find r along the left margin of the table

105. Using the Poisson Table Page A17
Recall, λ = 4
r = 0
r = 4
r = 7

106. Poisson Approximation to the Binomial

107. 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 If you play this game once every day, find the probability of winning exactly once in 365 days.
Solution: Consider a binomial experiment:
1. 𝑛=365
2. Because there is one winning set of numbers among the 10,000 that are possible
𝑝= 1 10,000

108. Example: Poisson Approximation to the Binomial
2. 𝑛𝑝< [ np = ]
Conclusion: We may use the Poisson distribution as an approximation to the binomial distribution. Winning is a “rare” event.
λ= 𝜇 = 𝑛𝑝 = 365 ⋅ 1 10,000 =0.0365
On average, we can expect to win times in a 365-day interval.

109. Example: Poisson Approximation to the Binomial
𝑃 1 = 𝑒 − ∙ !
= 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓 ,1 =0.035
Interpretive Statement: The probability of winning exactly once in a 365-day interval is about

110. Section 5.4 Summary
Found probabilities using the geometric distribution
Found probabilities using the Poisson distribution
Used a Poisson Approximation to the Binomial Distribution
Larson/Farber 4th ed