chapter 8

Other Probability Distributions

The last chapter explained the concepts of the binomial distribution. There are many other types of commonly used discrete distributions. A few are the multinomial distribution, the hypergeometric distribution, the Poisson distribution, and the geometric distribution. This chapter briefly explains the basic concepts of these distributions.

CHAPTER OBJECTIVES

In this chapter, you will learn

•  How to find the probability of an event using a multinomial distribution

•  How to find the probability of an event using a hypergeometric distribution

•  How to find the probability of an event using a geometric distribution

•  How to find the probability of an event using a Poisson distribution

The Multinomial Distribution

Recall that for a probability experiment to be a binomial experiment, two outcomes are necessary. But if each trial of a probability experiment has more than two outcomes, a distribution that can be used to describe the experiment is called a multinomial distribution. In addition, there must be a fixed number of independent trials, and the probability for success must remain the same for each trial.

A short version of the multinomial formula for three outcomes is given next. If X consists of mutually exclusive events E₁, E₂, and E₃, which have corresponding probabilities of p₁, p₂, and p₃ of occurring, and if x₁ is the number of times E₁ will occur, x₂ is the number of times E₂ will occur, and x₃ is the number of times E₃ will occur, then in n repeated trials, the probability of X is

 

_EXAMPLE

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A box contains four red balls, one blue ball, and five white balls. If five balls are selected with replacement, find the probability of getting one red ball, two blue balls, and two white balls.

 

SOLUTION

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n = 5, X₁ = 1, x₂ = 2, = x₃ = ₂, and p₁, p₂ and p₃. Hence, the probability of getting one red ball, two blue balls, and two white balls is

_EXAMPLE

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The probability that a restaurant will have no health violations, one health violation, or two health violations are 52%, 38%, and 10%, respectively. If eight restaurants are randomly selected, find the probability that three will have one violation and each of the two will have two violations.

 

SOLUTION

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n = 8, X₁ = 3, x₂ =3, x₃ = 2 and p₁ = 0.52, p₂ = 0.38, and p₃ = 0.10.

PRACTICE

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1.  In a survey of adults who purchased books at a book store, 60% chose a novel, 30% chose nonfiction, and 10% chose a self-help book. If 10 people are selected, find the probability that five selected a novel, three a nonfiction, and two a self-help book.

2.  At a fast-food restaurant, the probabilities that a person buys one item, two items, or three items are 0.2, 0.5, and 0.3. If 8 people are selected at random, find the probability that two will buy one item, four will buy two items, and two will buy three items.

3.  Motorcycles are randomly inspected in a certain state. The probabilities for having no violations, one violation, and two or more violations are 0.65, 0.28, and 0.07, respectively. If 10 motorcycles are inspected, find the probability that six will have no violations, three will have one violation, and one will have 2 or more violations.

4.  A die is rolled six times. Find the probability of getting 2 threes, 2 fours, and 2 sixes.

5.  According to Mendel’s theory, if tall and colorful plants are crossed with short and colorless plants, the corresponding probabilities are , , , and for tall and colorful, tall and colorless, short and colorful, and short and colorless plants, respectively. If 10 plants are selected, find the probability that five will be tall and colorful, three will be tall and colorless, one will be short and colorful, and one will be short and colorless.

 

ANSWERS

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1.  n = 10, x₁ = 5, x₂ = 3, x₃ = 2, and p₁ = 0.60, p₂ = 0.3, and p₃ = 0.1.

The probability is (rounded).

2.  n = 8, x₁ = 2, x₂ = 4, x₃ = 2, and p₁ = 0.2, p₂ = 0.5, and p₃ = 0.3.

The probability is

3.  n = 10, x₁ = 6, x₂ = 3, x₃ = 1, and p₁ = 0.65, p₂ =0.28, and p₃ = 0.07.

The probability is (rounded). .

4.  n = 6, x₁ = 2, x₂ = 2, x₃ = 2, and p₁ = , p₂ = , and p₃ = .

The probability is (rounded).

5.  n = 10, x₁ = 5, x₂ = 3, x₃ = 1, x₄ = 1, and p₁ =, p₂ =, p₃ =, and p₄ = .

The probability is (rounded).

The Hypergeometric Distribution

When a probability experiment has two outcomes and the items are selected without replacement, the hypergeometric distribution can be used to compute the probabilities. When there are two groups of items such that there are a items in the first group and b items in the second group, so that the total number of items is a + b, the probability of selecting x items from the first group and n – x items from the second group is

where n is the total number of items selected without replacement.

 

_EXAMPLE

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Six packets of seeds are randomly selected without replacement from a box that contains eight celery seed packets and four radish seed packets. Find the probability that the packer selects four celery seed packets and two radish seed packets.

 

SOLUTION

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Since there are eight celery seed packets and four radish seed packets, a = 8 and b = 4. The total number of selected packets is n = 4 + 2 = 6.We are looking for the probability of selecting four celery seed packets and two radish seed packets, so x = 4 and n – x = 6 – 4 = 2. The probability is

_EXAMPLE

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A budget clothing store has nine men’s suits. Five are black and four are gray. If four suits are sold, find the probability that three are black and one is gray.

 

SOLUTION

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Since there are five black suits and four gray suits, a = 5 and b = 4. If four suits are randomly sold and we want to know the probability that three are black and one is gray, n = 4, x = 3, and n – x = 4 – 3 = 1. The probability then is

_PRACTICE

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1.  A lot of 10 fire extinguishers contains three defective ones. If three are selected at random, find the probability that none are defective.

2.  A committee of five people is selected at random from four men and four women. Find the probability that two men and three women will be selected.

3.  There are 15 automobiles at a rental agency. Eight are white and seven are red. If six are rented at random, find the probability that three are white and three are red.

4.  If five cards are drawn at random from a deck of 52 cards, find the probability that exactly two are red cards.

5.  A freezer at a convenience store contains eight quarts of vanilla ice cream and six quarts of chocolate ice cream. If four quarts of ice cream are selected at random, find the probability that three are chocolate.

 

ANSWERS

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1.  a = 7, b = 3, n = 3, x = 0, n – x = 3.

2.  a = 4, b = 4, n = 5, x = 2, n – x = 3.

3.  a = 8, b = 7, n = 6, x = 3, n – x = 3.

4.  a = 26, b = 26, n = 5, x = 2, n – x = 3.

5.  a = 8, b = 6, n = 4, x = 3, n – x = 1.

The Geometric Distribution

Suppose you flip a coin several times. What is the probability that the first head appears on the third toss? In order to answer this question and other similar probability questions, the geometric distribution can be used. The formula for the probability that the first success occurs on the nth trial is

 

(1 – p)^{n – 1}p

 

where p is the probability of a success and n is the trial of the first success.

 

_EXAMPLE

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A coin is tossed. Find the probability that the first head occurs on the fourth toss.

 

SOLUTION

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The outcome is TTTH; hence, n = 4 and p so the probability of getting three tails and then a head is .

Using the formula given above,

_EXAMPLE

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A die is rolled. Find the probability of getting the first 3 on the third roll.

 

SOLUTION

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Let p and n = 3. hence,

The geometric distribution can be used to answer the question, “How long on average will I have to wait for a success?”

Suppose a person rolls a die until a 5 is obtained. The 5 could occur on the first roll (if one is lucky), on the second roll, on the third roll, etc. Now the question is, “On average, how many rolls would it take to get the first 5?” The answer is that if the probability of a success is p, then the average or expected number of independent trials it would take to get a success is In the dice situation, it would take on average 1 or six trials to get a 5. This is not so hard to believe since a 5 would occur on average one time in every six rolls because the probability of getting a 5 is .

 

_EXAMPLE

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A coin is tossed until a head is obtained. On average, how many trials would it take?

 

SOLUTION

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Since the probability of getting a head is , it would take 1 ÷ p trials.

On average it would take two trials.

Now suppose we ask, “If I roll a die, on average, how many trials would it take to get two 5s?” In this case, one five would occur on average once in the next six trials, so the second five would occur on average once in the next six trials. In general we would expect k successes on average in k/p trials. In this case, it would take 12 rolls.

 

_EXAMPLE

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If cards are selected from a deck and replaced, how many trials would it take on average to get three clubs?

 

SOLUTION

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Since there are 13 clubs in a deck of 52 cards, P(club) , The expected number of trials for selecting three clubs would be 12 draws.

This type of problem uses what is called the negative binomial distribution, which is a generalization of the geometric distribution.

Another interesting question one might ask is, “On average how many rolls of a die would it take to get all the faces, that is, 1 through 6?” In this case, the first roll would give 1 of the necessary numbers, so the probability of getting a number needed on the first roll would be 1. On the second roll, the probability of getting a number needed would be since there are five remaining needed numbers. The average number of rolls would be Since two numbers have been obtained, the probability of getting the next number would be . The average number of rolls would be . This would continue until all numbers are obtained. So the average number of rolls it would take to get all the numbers, 1 through 6, would be Hence, on average it would take about 14.7 or about 15 rolls to get all the numbers 1 through 6.

 

_EXAMPLE

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A children’s cereal manufacturer packages one toy spacecraft in each box. If there are five different spacecraft, and they are equally distributed, find the average number of boxes a child would have to purchase to get all five spacecrafts.

 

SOLUTION

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The probabilities are 1, , , , and , The average number of boxes for each are , , , , and ; so the total is which would mean a child on average would need to purchase 12 boxes of cereal since he or she cannot buy of a box.

_PRACTICE

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1.  A die is tossed until a 1, 2, 3, or 4 is obtained. Find the expected number of tosses.

2.  On average how many rolls of a die will it take to get four 5s?

3.  A card from an ordinary deck of cards is selected and then replaced. Another card is selected, etc. Find the probability that the first heart will occur on the fourth draw.

4.  A service station operator gives a scratch-off card with each fill-up over 8 gallons. On each card is one of six colors. When a customer gets all six colors, he wins 10 gallons of gasoline. Find the average number of fill-ups needed to win the 10 gallons.

5.  A coin is tossed until six heads are obtained. What is the expected number of tosses?

 

ANSWERS

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1.  p (round up to two, since 1.5 tosses are impossible).

2.

3.

4.

5.

The Poisson Distribution

Another commonly used discrete distribution is the Poisson distribution (named for Simeon D. Poisson, 1781–1840). This distribution is used when the variable occurs over a period of time, volume, area, etc. For example, it can be used to describe the arrivals of airplanes at an airport, the number of phone calls per hour for a 911 operator, the density of a certain species of plants over a geographic region, or the number of specific bacteria on a fixed surface.

The probability of x successes is

where e is a mathematical constant ≈2.7183 and λ is the mean or expected value of the variable.

 

NOTE The computations require a scientific calculator. Also, tables for values used in the Poisson distribution are available in some statistical textbooks.

 

_EXAMPLE

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A computer help hotline with a toll-free number receives an average of 5 calls per hour. For any given hour, find the probability that it will receive exactly 8 calls. Assume a Poisson distribution.

 

SOLUTION

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The mean, λ = 5 and x = 8. The probability is

Hence, there is about a 6.5% probability that the hotline will receive 8 calls.

_EXAMPLE

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If there are 100 typographical errors randomly distributed in a 600-page manuscript, find the probability that any given page has exactly two errors. Assume a Poisson distribution.

 

SOLUTION

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Find the mean numbers of errors: λ or 0.25. In other words, there is an average of 0.25 errors per page. In this case, x = 2, so the probability of selecting a page with exactly two errors is

Hence, the probability of two errors is about 2.4%.

_EXAMPLE

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Of a mail order company’s ads, 1.8% are returned because of the incorrect address or incomplete addresses. If the company sends 500 ads, find the probability that 6 will be returned. Assume a Poisson distribution.

 

SOLUTION

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The average is λ = 0.018 · 500 = 9 and x = 6.

The probability is about 9.1%.

_PRACTICE

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1.  A company that advertises on the radio in a certain area receives on average 11 calls every time it plays its radio commercials. Find the probability of getting 30 calls if the commercial is aired 3 times a day. Assume a Poisson distribution.

2.  In a batch of 50 calculators, on average, two are defective. In a random sample of 100 calculators, find the probability that 10 are defective. Assume a Poisson distribution.

3.  If there are three defects in every 1000 ft of cable, find the probability that there would be eight defects in 3000 ft of cable. Assume a Poisson distribution.

4.  A bus company finds, on average, that it has to pull out of service five buses per week. Find the probability that for a given week, seven buses will be pulled for repairs. Assume a Poisson distribution.

5.  The average number of calls per day to a poison control center is 14. On a given day, find the probability that the center will receive 12 calls.

 

ANSWERS

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1.  The average number of calls received is 11 per commercial airing, so if the commercial is aired three times, the average number of calls is 3 · 11 = 33 calls.

 λ = 33 and x = 30

2.  The average number of defective calculators is , so in a batch of 100 calculators, the average would be 100· (rounded).

 λ = 4 and x = 10

3.  The average number of defects in 1000 ft of cable is 3, so the average number of defects in 3000 ft of cable is .

 λ = 9 and x = 8

4.  In this case, λ = 5 and x = 7

5.  In this case, λ = 14 and x = 12

PROBABILITY SIDELIGHT: Some Thoughts about Lotteries

Today we are bombarded with lotteries. some people may have the fantasy of winning mega millions for a buck. each month the prizes seem to be getting larger and larger. each type of lottery gives the odds and the amount of the winnings. For some lotteries, the amount you win is based on the number of people who play. however, the more people who play the more chance there will be of multiple winners.

The odds and the expected value of a lottery game can be computed using combinations and the probability rules. For example, a lottery game in Pennsylvania is called “Match 6 Lotto.” For this game, a player selects 6 numbers from the numbers 1 to 49. If the player matches all six numbers, the player wins a minimum of $500,000. (Note: There are other ways of winning; and if there is no winner, the prize money is held over until the next drawing and increased a certain amount by the number of new players.) For now, just winning the $500,000 will be considered. In order to figure the odds, it is necessary to figure the number of winning combinations. In this case, we are selecting without regard to order six numbers from 49 numbers. Hence, there are ₄₉C₆ or 13,983,816 ways to select a ticket. However, the odds given in the lottery brochure are 1:4,661,272. The reason is that if you select six numbers, you can have the computer select two more sets of six numbers, giving you three chances to win. So dividing 13,983,816 by 3, you get 4,661,272, and the odds are 1:4,661,272.

Now let’s make some sense of this. There are 60 seconds in a minute, 60 minutes in an hour, and 24 hours in a day. So there are 60 × 60 × 24 = 86,400 seconds in a day. If you divide 4,661,272 by 86,400, you get approximately 54 days. So selecting a winner would be like selecting a given random second in a time period of 54 days!

What about a guaranteed method to win the lottery? It does exist. If you purchased all possible number combinations, then you would be assured of winning, wouldn’t you? But is it possible?

A group of investors from Melbourne, Australia, decided to try. At the time, they decided to attempt to win a $27 million prize given by the Virginia State Lottery. The lottery consisted of selecting 6 numbers out of a possible 44. This means that they would have to purchase ₄₄C₆ or 7,059,052 different tickets. At $1 a ticket, they would need to raise $7,059,052. However, the profit would be somewhere near $20 million if they won. Next there is always a possibility of having to split the winnings with other winners, thus reducing the profit. Finally, they need to buy the 7 or so million tickets within the 72-hour time frame. Understand that this group sent out teams, and they were able to purchase about 5 million tickets, and they did win the money without having to split the profit! Some objections were raised by the other players (i.e., losers), but the group was eventually paid off.

By the way, it does not matter which numbers you play on the lottery since the drawing is random and every combination has the same probability of occurring. Some people suggest unusual combinations such as 1, 2, 3, 4, 5, and 6 are better, since there is less of a chance of having to split your winnings if you do indeed win.

So what does this all mean? I heard a mathematician sum it all up by saying that you have the same chance of winning a big jackpot on a state lottery, whether or not you purchase a ticket.

 

Summary

There are many types of discrete probability distributions besides the binomial distribution. The most common ones are the multinomial distribution, the hypergeometric distribution, the geometric distribution, and the Poisson distribution.

The multinomial distribution is an extension of the binomial distribution and is used when there are three or more independent outcomes for a probability experiment.

The hypergeometric distribution is used when sampling is done without replacement. The geometric distribution is used to determine the probability of an outcome occurring on a specific trial; this distribution can also be used to find the probability of the first occurrence of an outcome.

The Poisson distribution is used when the variable occurs over a period of time, over a period of area or volume, etc.

In addition there are other discrete probability distributions used in mathematics; however, these are beyond the scope of this book.

 

QUIZ

 

1.  Which distribution can be used when there are three or more outcomes?

A.  Poisson

B.  hypergeometric

C.  Multinomial

D.  Geometric

 

2.  Which distribution requires that sampling be done without replacement?

A.  Poisson

B.  Geometric

C.  Multinomial

D.  hypergeometric

 

3.  Which distribution can be used to determine the probability of an outcome occurring on a specific trial?

A.  Geometric

B.  Poisson

C.  Multinomial

D.  hypergeometric

 

4.  Which distribution can be used when the variable occurs over time?

A.  Poisson

B.  hypergeometric

C.  Multinomial

D.  Geometric

 

5.  If 4 cards are drawn from a deck without replacement, the probability that exactly 2 hearts will be selected is

A.  0.035

B.  0.1875

C.  0.213

D.  0.211

 

6.  The probabilities that an income tax return will have 0, 1, 2, or 3 mathematical errors are 0.6, 0.2, 0.15, or 0.05, respectively. If 5 returns are randomly selected, the probability that 2 will contain no errors, 1 will contain one error, one will contain 2 errors, and one will contain 3 errors is

A.  0.0615

B.  0.0442

C.  0.0324

D.  0.0178

 

7.  A die is rolled five times. The probability of getting two 1s, two 4s, and one 6 is

A.  0.004

B.  0.006

C.  0.013

D.  0.025

 

8.  A college club consists of 15 women and 12 men. If a committee of six students is selected at random, the probability that exactly two students are men is

A.  0.0057

B.  0.3043

C.  0.2164

D.  0.0039

 

9.  Of the 12 surfboards in a surf shop, four are white. If five are selected at random to be placed outside on a given day, the probability that exactly two are white is

A.  0.037

B.  0.424

C.  0.22

D.  0.056

 

10.  About 8% of the cat population carries a certain genetic trait. Assume the distribution is Poisson. The probability that in a group of 200 randomly selected cats, 14 cats carry the gene is

A.  0.161

B.  0.093

C.  0.042

D.  0.159

 

11.  The number of boating accidents on a large lake follows a Poisson distribution. The probability of an accident is 0.02. If there are 600 boats on the lake on a summer day, the probability that there will be exactly 11 accidents will be

A.  0.167

B.  0.248

C.  0.127

D.  0.114

 

12.  When a coin is tossed, the probability of getting the first head on the third toss is

A.  

B.  

C.  

D.  

 

13.  Eight cards are numbered 1 through 8. The cards are mixed. A card is selected, and its number is recorded. Then it is replaced, and another one is selected and so on. On average how many cards will it take to get all the numbers once?

A.  28.62

B.  64

C.  21.74

D.  32

 

14.  Cards are selected from a deck of 52 cards and replaced. Find the average number of cards it will take before a face card is selected.

A.  12

B.  

C.  8

D.  

 

15.  A 10-sided die is rolled; the average number of tosses that it will take to get four 5s is

A.  30

B.  8

C.  18

D.  40