Binomial & Geometric Distribution Problems



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The Binomial Distribution

Characteristics of a Binomial Setting:

1. Each observation (or trial) has two categories: success or failure.

2. There is a fixed number of observations (n).

3. The observations are all independent.

4. The probability of success (p) is the same for each trial.

Binomial or Not?

For each of the following examples, we wish to determine if the situation described is binomial. To test this, we must check all four of the conditions listed above.

1. Blood type is inherited. If both parents carry genes for the O and A blood types, each child has probability 0.25 of getting two O genes and so of having blood type O. Different children inherit independently of each other. We wish to explore the number of children out of 5 children born to these parents will have blood type O.

1. Two Categories?

Yes, in this case type O blood = success and any other blood type = failure

2. Fixed number n?

Yes, n = 5 (since there are 5 children)

3. Independent Observations?

Yes, we are told children inherit blood types independently

4. p is the same?

Yes, for each child the probability of type O is the same (p = 0.25)

Since all 4 conditions are met, this distribution is BINOMIAL (

Binomial or Not? (continued from previous pg)

2. Deal 10 cards from a shuffled deck and count the number of red cards.

1. Two Categories?

Yes, in this case red card = success and black card = failure

2. Fixed number n?

Yes, n = 10 (since there are 10 cards)

3. Independent Observations?

NO, cards are not replaced – so they are not independent

4. p is the same?

NO, cards are not replaced – so p will change based as each new card is drawn.

Since all 4 conditions are NOT met, this distribution is NOT BINOMIAL

3. Engineers define reliability as the probability that an item will perform its function under specific conditions for a specific period of time. Suppose a flight engineer tells us that an aircraft engine turbine has probability 0.999 of performing properly for an hour of flight. We have a fleet of 350 engines that will fly for an hour. We wish to find the probability that all 350 engines perform without failure.

1. Two Categories?

Yes, in this case the engine working properly = success and engine failure = failure

2. Fixed number n?

Yes, n = 350 engines

3. Independent Observations?

Yes, safe to assume that one engine failing or working will not affect another

4. p is the same?

Yes, we are given that the probability of success is 0.999; so, p = 0.999

BINOMIAL DISTRIBUTION (

Your Turn: Binomial or Not?

Using the three previous examples as models, determine if each of the following scenarios describe binomial distributions or not. If the distribution is not binomial, explain why it is not. If it is binomial, you need to list and verify that the four conditions are met.

1. An auto manufacturer chooses one car from each hour’s production for a detailed quality inspection. One variable recorded is the count of finish defects (dimples, ripples, etc.) in the car’s paint.

Not Binomial

Does not have a set number of trials n, different types of defects would have different probabilities.

2. The pool of potential jurors for a murder case contains 100 persons chosen at random form the adult residents of a large city. Each person in the pool is asked whether he or she opposes the death penalty. We are interested in the number who say “yes.”

Binomial, if we know that p is the same for every random person

n = 100

Success = Person opposes death penalty

Failure = Person supports death penalty

It is reasonable to assume each person’s answer would be independent

3. When an opinion poll calls residential telephone numbers at random, only 20% of the calls reach a live person. You watch the random dialing machine make 15 calls. We are interested in the number of times the machine reaches a live person.

Binomial

n = 15

Success = reaching live person & failure = reaching a machine

p = 0.20

Trials are independent

Dir: Determine if each of the following scenarios describe binomial distributions or not. If the distribution is not binomial, explain why it is not. If it is binomial, you need to list and verify that the four conditions are met.

4. At peak periods, 15% of attempted log-ins to an email service fail. Log-in attempts are independent and each has the same probability of failing. Darci logs in repeatedly until she succeeds. We are interested in the number of attempts she makes before successfully logging into her account.

Not Binomial since we do not have a set n (this example is geometric)

5. A student studies binomial distributions using computer-assisted instruction. After the lesson, the computer presents 10 problems. The student solves each problem and enters her answer. The computer gives additional instruction between problems if the answer is wrong. We are interested in the number of problems that the students answers correctly.

Not Binomial

Independence fails since the computer provides additional instruction

Finding Binomial Probabilities in the TI-83/TI-84

Binompdf:

Located under normalcdf and invnorm (2nd Vars)

Used when we want a specific number of successes.

Used with problems of the type P(x = 3).

“pdf” = probability distribution function.

Binompdf(n, p, r)

• Select binompdf from the calculator list.

• In the open parenthesis, we enter n, p, r.

n = number of trials/observations

p = probability of success

r = desired number of successes

Example:

Among employed women, 25% have never been married. Select 10 employed women at random.

1. What is the probability that exactly 2 women in my sample have never been married?

2. What is the probability that 5 of the women in the sample have never been married?

Notes & Answers:

We have a binomial distribution with n = 10 and p = .25.

In #1, we wish to find the probability of exactly 2 successes, so r = 2.

Use the calculator to enter:

Binompdf(10, 0.25, 2)

Our Answer is: 0.2816

In #2, we want the number of successes to be 5 (r = 5).

Use the calculator to enter

Binompdf(10, 0.25, 5)

Our Answer is: 0.0584

Binomial “pdf” Practice Problems:

Use the binomial pdf function in your graphing calculator to answer each of the following questions. Be sure to included “binompdf(n, p, r)” with the appropriate numbers for n, p, and r with your answer.

1. The count of children with type O blood among 5 children whose parents carry genes for both O and the A blood type follows a binomial distribution with n = 5 and p = 0.25. Find the probability that exactly 3 children have type O blood.

Binompfd(5, 0.25, 3) = 0.0879

2. Suppose you purchase a bundle of 10 bare-root broccoli plants. The sales clerk tells you that on average you can expect 5% of the plants to die before producing any broccoli. Assume that the bundle is a random sample of plants.

a. Find the probability that you will lose one plant before it produces broccoli.

Note: let X = number of broccoli plants that you lose

We are looking for P(x = 1)

Binompdf(10, .05, 1) = 0.315

b. Find the probability that you will not lose any plants before they produce broccoli. (This means we want r = 0.)

Binompdf(10, .05, 0) = 0.587

c. Add the two probabilities found in parts a and b. What does this probability represent?

0.315 + 0.587 = 0.914

This is the probability of one of less of the plants dying. P(x < 1)

3. A university claims that 80% of its basketball players get degrees. An investigation examines the fate of all 20 players who entered the program over a period of several years. Of these players, 11 graduated and the remaining 9 are no longer in school. If the university’s claim is true, the number of players among the 20 who graduate should have the binomial distribution with n = 20 and p = 0.8. Find the probability that….

a. exactly 11 out of the 20 players graduate.

Binompdf(20, .8, 11) = 0.0074

b. all 20 of the players graduate.

Binompdf(20, .8, 20) = 0.0115

c. use your answer from part b to find the probability that not all 20 players graduate.

1 - .0115 = 0.9985

4. Suppose that James guesses on each question of a 50-item true-false quiz.

a. Assuming that James guesses are independent of each other, are the conditions for a binomial distribution satisfied? If so, what are p and n?

Yes: n = 50, p = 0.5, success = correct answer & failure = incorrect answer, trials are independent

b. Find the probability that James answers 20 of the questions correctly.

Binompdf(50, 0.5, 20) = 0.419

c. Find the probability that James answers 15 questions correctly.

Binompdf(50, 0.5, 15) = 0.002

Binomcdf:

Use when we want less than or equal to a specific number of successes:

Problems of the type P(x < 3)

“cdf” = cumulative distribution function

Binomcdf(n, p, r)

• Select binomcdf from the calculator list.

• In the open parenthesis, we enter n, p, r.

n = number of trials/observations

p = probability of success

r = desired number of successes

Important Note:

Binomcdf only finds the probability that the number of successes is less than or equal to the desired value. It always finds P(x < r).

• Be Careful in wording your problems.

• If you want the probability that the number of successes is less than 2, you must

find P(x < 1).

Example:

Among employed women, 25% have never been married. Select 10 employed women at random.

1. What is the probability that fewer than 5 women have never been married?

2. What is the probability that at most 2 women in the sample have never been married?

Notes & Answers:

We have a binomial distribution with n = 10 and p = .25.

In #1, we wish to find the prob. that the number of successes is less than 5, i.e., P(x < 4.)

Use the calculator to enter:

Binomcdf(10, 0.25, 4)

Our Answer is: 0.9219

In #2, we want the number of successes to be 2 or less, P(x < 2).

Use the calculator to enter:

Binomcdf(10, 0.25, 2)

Our Answer is: 0.5256

Binomial “cdf” Practice Problems:

1. The count of children with type O blood among 5 children whose parents carry genes for both O and the A blood type follows a binomial distribution with n = 5 and p = 0.25. Find the probability that at most 3 of the children have type O blood.

P(x < 3) = binomcdf(5, 0.35, 3) = 0.946

5. Suppose you purchase a bundle of 10 bare-root broccoli plants. The sales clerk tells you that on average you can expect 5% of the plants to die before producing any broccoli. Assume that the bundle is a random sample of plants.

a. Find the probability that you will lose at most one plant before it produces broccoli.

P(x < 1) = binomcdf(10, 0.5, 1) = 0.9139

b. Find the probability that you will lose fewer than 3 plants before they produce broccoli.

P(x < 3) = binomcdf(10, 0.5, 2) = 0.9885

6. A university claims that 80% of its basketball players get degrees. An investigation examines the fate of all 20 players who entered the program over a period of several years. Of these players, 11 graduated and the remaining 9 are no longer in school. If the university’s claim is true, the number of players among the 20 who graduate should have the binomial distribution with n = 20 and p = 0.8.

a. Find the probability that ten or fewer of the players graduate.

P(x < 10) = binomcdf(20, 0.8, 10) = 0.0026

b. Use your answer to part a, to determine the probability that more than 10 graduate. (Note: these two events are complementary.

1 – 0.0026 = 0.9974

c. Find the probability that at most 11 graduate.

P(x < 11) = binomcdf(20, 0.8, 11) = 0.00998

7. Suppose that James guesses on each question of a 50-item true-false quiz.

a. Find the probability that James answers at most 20 of the questions correctly.

P(x < 20) = binomcdf(50, 0.5, 20) = 0.1013

b. Find the probability that James answers 25 or fewer of the questions correctly.

P(x < 25) = binomcdf(50, .5, 25) = 0.5561

Mixing it Up:

The following problems review all that you have learned in this packet. The problems are mixed together, so you will need to think about them before plugging numbers into your calculator. (

1. According to a 2000 study by the Bureau of Justice Statistics, approximately 2% of the nation’s 72 million children had a parent behind bars – nearly 1.5 million minors. Suppose that 100 children are randomly selected. Let X be the number of children who have an incarcerated parent.

a. Does X satisfy the requirements for a binomial setting? If so, what are n and p?

Yes, n = 100, p = .02, success = incarcerated parent, trials are independent

b. Describe P(X = 0) in words.

This is the probability that no child has an incarcerated parent

c. Find P(X = 0).

Binompdf(100, 0.2. 0) = 0.0000000002 or ~0

d. Find P(X = 1).

Binompdf(100, 0.2. 1) = 0.000000005 or ~0

e. Use your answers to c and d, to find the probability that 2 or more of the 100 children have a parent behind bars. (Caution: binomcdf will not work, since I want more than 2….. You must use the answers to c and d to find this new probability. Tricky….)

~100% (Find exact probability by subtracting answers to c & d from 1)

2. Suppose that Erin guesses on each question of a multiple-choice quiz. If each question has four different choices, find the probability that Erin gets at most 4 questions correct on a ten-question quiz.

P(x < 4) = binomcdf(10, 0.25, 4) = 0.9218

3. Suppose that Erin guesses on each question of a multiple-choice quiz. If each question has five different choices, find the probability that Erin gets exactly five correct on a ten-question quiz.

P(x = 5) – binompdf(10, 0.2, 5) = 0.0264

4. Suppose that Erin guesses on each question of a multiple-choice quiz. If each question has four choices and there are a total of 12 questions, find the probability that Erin…

a. Gets exactly one question correct.

P(x = 1) = binompdf(12, 0.25, 1) = 0.1267

b. Does not get any questions correct.

P(x = 0) = binompdf(12, 0.25, 0) = 0.0317

c. Use your answer/answers above, to find the probability that Erin gets 1 or more correct answers on this quiz.

1 – 0.1267 – 0.0317 = 0.8416

5. In a test for ESP (extrasensory perception), a subject is told that cards the experimenter can see but the subject cannot contain a star, a circle, a wave, or a square. As the experimenter looks at each of 20 cards in turn, the subject names the shape on the card. A subject who is just guessing has probability 0.25 of guessing correctly on each card.

a. The count of guesses in 20 cards has a binomial distribution. What are n and p?

n = 20 & p = 0.25

b. What is the probability of exactly 5 correct guesses?

P(x = 5) = binompdf(20, .25, 5) = 0.2023

c. What is the probability of at least 5 correct guesses?

P(x > 5) = 1 – binomcdf(20, .25, 4) = 0.5852

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