Chapter Five The Binomial Probability Distribution and Related … · 2016. 2. 6. · Probability...

Chapter Five

The Binomial Probability Distribution

and Related Topics

Section 4

The Geometric and Poisson Probability

Distributions

Essential Questions

How are Geometric and Poisson probability distributions different from the binomial probability distribution? How are they the same?

Student ObjectivesThe student will determine the probability of an event using the geometric probability distribution.The student will determine the probability of an event using the Poisson probability distribution.The student will use the Poisson probability distribution to approximate the probability of a binomial experiment.The student will determine the probability of an event using the appropriate commands on their calculator.The student will determine the probability of an event using the appropriate formula.The student will determine the probability of an event using the appropriate chart.

Key Terms

Geometric probability distribution

Negative binomial distribution

Poisson probability distribution

Geometric Probability Distribution

Geomeric Probability Distribution

P n( ) = p 1− p( )n−1

where n is the number of binomial trials on which the first success occurs (n = 1, 2, 3, ...) and p is the probability of success on each trial. Notice: p is the same for each trial. Using some mathematics involving infinite series, it can be shown that the population mean and standard deviation of the geometric distribution are:

µ= 1p

and σ =1− pp


Geomeric Probability Distribution

P n( ) = p 1− p( )n−1

Calculator commands Single probability 2nd DISTR (Option E) P r( ) = geometpdf p, r( )or Cummulative probability ( 0, 1, 2, 3, ..., r( ) 2nd DISTR (Option F) P r( ) = geometcdf p, r( )

A student must come to the front of the room and correctly answer a randomly chosen question before returning to their seat. The probability of answering the questions has been determined to be 0.75.


(a) What is the mean number of questions that will need to be answered to sit down?

(b) What is the standard deviation? (c) What is the probability of having to answer exactly 3

questions before getting to sit down? (Formula and Calculator)

(d) What is the probability of being able to sit down by answering at most 5 questions? (Calculator)

A student must come to the front of the room and correctly answer a randomly chosen question before returning to their seat. The probability of answering the questions has been determined to be


(a) What is the mean number of questions that will need to be answered to sit down?

µ = 1p

µ = 10.75

µ = 1.3333



(b) What is the standard deviation?σ =

qp

σ = 0.250.75

σ = 0.50.75

σ = 0.6667



(c) What is the probability of having to answer exactly 3 questions before getting to sit down? (Formula and Calculator)P r( ) = p q( )r−1

P 3( )0.75( ) 0.25( )2

0.75( ) 0.0625( )0.04688

P r( ) = geometpdf p, r( )P 3( )

geometpdf 0.75, 3( )0.04688



(d) What is the probability of being able to sit down by answering at most 5 questions?? (Calculator)

P r ≤ k( ) = geometcdf p, k( )P r ≤ 5( )

geometcdf 0.75, 5( )0.9990

Sample QuestionYou are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer.

(a) What is the probability of winning the first time you play the game?

(b) The second time? (c) The third time? (d) How many games would you need to play to be 95% certain

that you would win?


(a) What is the probability of winning the first time you play the game?

1st gameP n = 1( )13

⎛⎝⎜

⎞⎠⎟

23

⎛⎝⎜

⎞⎠⎟

0

13

⎛⎝⎜

⎞⎠⎟ 1( )

13

0.3333


(b) The second time? 2nd gameP n = 2( )13

⎛⎝⎜

⎞⎠⎟

23

⎛⎝⎜

⎞⎠⎟

1

13

⎛⎝⎜

⎞⎠⎟

23

⎛⎝⎜

⎞⎠⎟

29

0.2222

Sample QuestionYou are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. 3rd game

P n = 3( )13

⎛⎝⎜

⎞⎠⎟

23

⎛⎝⎜

⎞⎠⎟

2

13

⎛⎝⎜

⎞⎠⎟

49

⎛⎝⎜

⎞⎠⎟

427

0.1481

(c) The third time?


(d) How many games would you need to play to be 95% certain that you would win?

P n ≤ 8( )geometcdf 1/ 3, 8( )

0.9610


(a) What is the mean number of games required to win the game?

(b) What is the standard deviation?


(a) What is the mean number of games required to win the game?

µ = 1p

µ = 113

⎛⎝⎜

⎞⎠⎟

µ = 3


(b) What is the standard deviation?σ =

qp

σ =

2313

σ = 0.66670.3333

σ = 0.81650.3333

σ = 2.4495

Poisson Probability Distribution

Poisson Probability DistributionLet λ (Greek letter lambda) be the mean number of success over time, volume, area, and so forth. Let r, be the number of successes (r = 0, 1, 2, 3,...) in a corresponding interval of time, volume, area, andso forth. Then the probability of r successes in the interval is

P r( ) = e−λλ r

r!where e is approximately equal to 2.7183. Using some mathematics involving infinite series, it can be shown that the population mean and standard deviation of the Poisson distribution are:

µ = λ and σ = λ



P r = k( ) = e−λ λ( )kk!

Calculator commands Single probability 2nd DISTR (Option C) P r = k( ) = poissonpdf λ, k( )or Cummulative probability 0, 1, 2, 3, ..., k( ) 2nd DISTR (Option D) P r ≤ k( ) = poissoncdf p, k( )


The typical teenage sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days.(a) What is the mean number of text messages? (b) What is the standard deviation? (c) What is the probability of sending 10 text messages during this time

period? (Formula) (d) What is the probability of sending 15 text messages during this time

period? (Calculator) (e) What is the probability of sending 12 text messages during this time

period? (Chart - Round the value of lambda to the nearest 10th) (f) What is the probability of sending at least 5 text messages?

(Calculator)


The typical teenage sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days.

(a) What is the mean number of text messages?

λ = 3339 messages1 month

× 1 month30.5 days

× 1 day24 hours

× 2 hours2 hours

λ = 9.1230 messages2-hours

λ = 9.1messages2-hours



(b) What is the standard deviation?

σ = λ

σ = 9.1230σ = 3.0204



(c) What is the probability of sending 10 text messages during this time period? (Formula)

P r = 10( )e−9.1230( ) 9.1230( )10

10!0.0001091( ) 3993496762( )

3628800435818.97736288000.1201



(d) What is the probability of sending 15 text messages during this time period? (Calculator) P r = 15( )

poissonpdf 9.1230, 15( )0.02106



(e) What is the probability of sending 12 text messages during this time period? (Chart - Round the value of lambda to the nearest 10th)

λ = 9.1P r = 12( )0.0752


The typical teenage sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days.(f) What is the probability of

sending at least 5 text messages? (Calculator)

P r ≥ 5( )1− P r ≤ 4( )

1− poissoncdf 9.1230, 4( )1− 0.0510

0.9490

Sample QuestionIt has been determined that about that 14 cars run a red-light in downtown Annville every day. The police have decided to place a patrol car in the area for a 10-hour period.

(a) Round your value of lambda to the nearest tenth and use the chart to determine the probability of 0, 1, and 2 cars running the red light.

(b) Use the formula and the exact value of lambda to determine the probability of 5 cars running the red light.

(c) Use the calculator to determine the probability of 4 cars running the red light.

(d) Use the calculator to determine the probability that more than 7 cars will run the red light.


(a) Round your value of lambda to the nearest tenth and use the chart to determine the probability of 0, 1, and 2 cars running the red light.

λ = 14 cars1 day

× 1 day24 hours

× 10 hours1 shift

λ = 5.8333 carsshift

λ = 5.8 carsshift

P r = 0( )0.0030

P r = 1( )0.0176

P r = 2( )0.0509


(b) Use the formula and the exact value of lambda to determine the probability of 5 cars running the red light.

P r = 5( )e−5.8333( ) 5.8333( )5

5!0.0029( ) 6754.3564( )

12019.77881200.1648


(c) Use the calculator to determine the probability of 4 cars running the red light. P r = 4( )

poissonpdf 5.8333, 4( )0.1413


(d) Use the calculator to determine the probability that 7 or more cars will run the red light.

P r ≥ 7( )1− P r ≤ 6( )

1− poissoncdf 5.8333, 6( )1− 0.6331

0.3669

Binomial vs PoissonHow to approximate Binomial Probabilities using Poisson ProbabilitiesSuppose you have a binomial distribution with n = number of trials p = probability of succes in each trial, and r = the number of successIf n ≥100 and np <10, then r has a binomial distribution that is approximated by a Poisson distribution with λ = np.

P r( ) = e−λλ r

r!Note: λ = np is the expected value of the binomial distribution.

Binomial vs PoissonThe population of Cleona borough is approximated at 2,091. Roughly 0.29% of the population owns some type of exotic wildlife.(a) Why is the Poisson probability distribution a good

approximation of the binomial distribution? (b) What is the value of lambda? (c) What is the probability that 2 people will own some exotic

wildlife? (Formula - Binomial and Poisson) (d) What is the probability of having at least 7 people owning some

type of exotic wildlife? (Calculator - Binomial and Poisson)

Binomial vs PoissonThe population of Cleona borough is approximated at 2,091. Roughly 0.29% of the population owns some type of exotic wildlife.

(a) Why is the Poisson probability distribution a good approximation of the binomial distribution?

n = 2091Since n ≥ 100, the population is large enough touse the Poisson distribution to approximate the binomial probability.

np = 2091 0.0029( )np = 6.0639Since np < 10, the expected value of the binomial distribution is small enough to use the Poisson distribution to approximate the binomial probability.

Yes, the Poisson distribution is a good approximation of the binomial

distribution since n ≥ 100 and np < 10.


(b) What is the value of lambda?

np = 2091 0.0029( )np = 6.0639λ = 6.0639


(c) What is the probability that 2 people will own some exotic wildlife? (Formula - Binomial and Poisson)

Binomial vs. PoissonP r = 2( ) P r = 2( )

2091C2 0.0029( )2 0.9971( )2089 e−6.0639( ) 6.0639( )2

2!2091( ) 2090( )

2( ) 1( )⎛⎝⎜

⎞⎠⎟0.0029( )2 0.9971( )2089 0.002325( ) 36.7709( )

2

43701902

⎛⎝⎜

⎞⎠⎟ 0.0029( )2 0.9971( )2089 0.08550

22185095( ) 0.000008410( ) 0.002318( ) 0.04275

0.04260


(d) What is the probability of having at least 7 people owning some type of exotic wildlife? (Calculator - Binomial and Poisson)

Binomial vs. PoissonP r ≥ 7( ) P r ≥ 7( )

1− P r ≤ 6( ) 1− P r ≤ 6( )1− binomcdf 2091, 0.0029, 6( ) 1− poissoncdf 6.0639, 6( )

1− 0.5960 1− 0.59600.4040 0.4040

Sample QuestionIn a recent study of the high school graduates it has been determined that the probability that a graduate will be in serious trouble with the law within 5 years of their graduation is 0.025. There are 144 students in the graduating class.(a) Is the Poisson distribution a good approximation of the binomial

distribution? Explain. (b) Use the formula for binomial probability to determine the

probability that 4 students will get in serious trouble with the law within 5 years of their graduation. Repeat with the Poisson formula.

(c) Use your calculator to determine the probability that at least 10 students will get in serious trouble with the law. Repeat with the Poisson formulas. Are the answers similar?

Sample QuestionIn a recent study of the high school graduates it has been determined that the probability that a graduate will be in serious trouble with the law within 5 years of their graduation is 0.025. There are 144 students in the graduating class.

(a) Is the Poisson distribution a good approximation of the binomial distribution? Explain.

n = 144Since n ≥ 100, the population is large enough touse the Poisson distribution to approximate the binomial probability.

np = 144 0.025( )np = 3.6Since np < 10, the expected value of the binomial distribution is small enough to use the Poisson distribution to approximate the binomial probability.

Yes, the Poisson distribution is a good approximation of the binomial distribution since n ≥ 100 and np < 10.


(b) Use the formula for binomial probability to determine the probability that 4 students will get in serious trouble with the law within 5 years of graduation. Repeat with the Poisson formula. P 4( )

144C4 0.025( )4 0.975( )140

144 ⋅143⋅142 ⋅1414 ⋅3⋅2 ⋅1

⎛⎝⎜

⎞⎠⎟ 0.025( )4 0.975( )140

41229302424

⎛⎝⎜

⎞⎠⎟ 0.025( )4 0.975( )140

17178876( ) 0.0000003906( ) 0.02888( )0.1938

P 4( )e−3.6 3.6( )4

4!0.02732( ) 167.9616( )

244.589324

0.1912


(c) Use your calculator to determine the probability that at least 10 students will get in serious trouble with the law. Repeat with the Poisson formulas. Are the answers similar?

P r ≥10( )1− P r ≤ 9( )

1− binomcdf 144, 0.0250, 9( )1− 0.99650.003522

P r ≥10( )1− P r ≤ 9( )

1− poissoncdf 3.6, 9( )1− 0.99600.004024

Yes, the answers only differ by 0.0005021.

Negative Binomial Probability Distribution

Negative Binomial DistributionLet k ≥1 be a fixed whole number. The probability that the kth success occurs on trial n is

P n( ) = n−1Ck−1 p( )k q( )n−k

where

n−1Ck−1 =n −1( )!

k −1( )! n − k( )! n = k, k +1, k + 2, ...The expected value and standard deviation of the geometric distribution are:

µ = kp

and σ =kqp

Note: if k = 1, the negative binomial distribution is called the geometric distribution.


Dylan can leave the final examination when he is finished with the examination. Dylan decided that he needed only to answer 5 questions correct in order to pass the course. There were a total of 100 questions on the final examination. The probability that he can correctly answer any of the questions is 0.65.

(a) What is the mean number of questions that will need to be answered to pass the course?

(b) What is the standard deviation? (c) What is the probability that he answers 5 correct in the

first 10 questions?



(a) What is the mean number of questions that will need to be answered to pass the course?

µ = 50.65

µ = 7.6923



(b) What is the standard deviation? σ =5( ) 0.35( )0.65

σ = 1.750.65

σ = 1.32290.65

σ = 2.0352



(c) What is the probability that he answers 5 correct in the first 10 questions?

P 5 out of 10( )9C4 0.65( )5 0.35( )5

9( ) 8( ) 7( ) 6( )4( ) 3( ) 2( ) 1( )

⎛⎝⎜

⎞⎠⎟

0.65( )5 0.35( )5

302424

⎛⎝⎜

⎞⎠⎟ 0.65( )5 0.35( )5

126( ) 0.1160( ) 0.005252( )0.07679

Negative Binomial Probability and Geometric Distribution

Negative Binomial Distribution vs Geometric Distributionk ⇐ number of successes ⇒ 1

k −1 ⇐successes priorto the last trial

⇒ 0

n ⇐ number of trials ⇒ np ⇐ probability of a success ⇒ pkp

⇐ µ ⇒ 1p

kqp

⇐ σ ⇒qp

n−1Ck−1pkqn−k ⇐ Probability formula ⇒ p1qn−1

The geometric distribution has only one success and it must occur on the last trial. The negative binomial distribution may have multiple successes but the last trial must be a success.

Chapter Five The Binomial Probability Distribution and Related … · 2016. 2. 6. · Probability...

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