1. Binomial & Geometric Random Variables
Section 6.3
Reference Text:
The Practice of Statistics, Fourth Edition.
Starnes, Yates, Moore
Lesson 6.1.1

2. Objectives Binomial Random Variables and Binomial Distribution
Requirements to be Binomial- B.I.N.S
Binomial Coefficient: by formula and TI-83
Binomial Probability: by formula and TI-83
Mean and Standard Deviation
“10% condition” sampling w/o replacement
Geometric Random Variables and Geometric Distribution
Requirements to be Geometric- B.I.T.S
Geometric Probability: by formula and TI-83
Mean (expected value)

3. Consider this: Toss a coin 5 times, count the number of heads.
Spin a roulette wheel 8 times. Record how many times a ball lands in a red slot.
Take a random sample of 100 babies born in U.S hospitals today. Count the number of females.
In each case, we’re preforming repeated trials of the same chance process. The number of trials is fixed in advance. (number of trials n) In addition, the outcome of one trial has no effect on the outcome of any other trial, that is, the trials are independent. We’re interested in the number of times a specific event occurs, (we’ll call it a “success”). Our chances of getting a “success” are the same on each trial. (probability p is the same)

4. Binomial and Geometric Random Variables
Definition:
A binomial setting arises when we perform several independent trials of the same chance process and record the number of times that a particular outcome occurs. The four conditions for a binomial setting are
Binomial and Geometric Random Variables
• Binary? The possible outcomes of each trial can be classified as “success” or “failure.” B
• Independent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial. I
• Number? The number of trials n of the chance process must be fixed in advance. N S
• Success? On each trial, the probability p of success must be the same.

5. Consider the following 3 scenarios
In each case, determine whether the given random variable has a binomial distribution.
Justify your answer. B.I.N.S
Genetics say that children receive genes from each of their parents independently. Each child of a particular pair of parents have a probability 0.25 of having type O blood. Suppose these parents have 5 children. Let 𝑋 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐ℎ𝑖𝑙𝑑𝑟𝑒𝑛 𝑤𝑖𝑡ℎ 𝑡𝑦𝑝𝑒 𝑂.
Binary? “success” = has type O blood, “failure” = does not have type O
Independent? – children inherit genes are independent of each parent.
Number? There are 𝑛=5 number of trials
Success? The probability of a success is 𝑝=0.25
This is a binomial setting, 𝑛 = 5 𝑎𝑛𝑑 𝑝 = 0.25 B I N S

6. Consider the following 3 scenarios
2. Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y= the number of Aces you observe.
Binary?
“success” = is an ace, “failure” = is not an ace
Independent?
If the first card that gets turned over is an Ace then there is less of a probability to get another ace, not independent
This is not a binomial setting since they are not independent. B I N S

7. Consider the following 3 scenarios
3. Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and shuffle again. Repeat this process until you get an ace. Let W= the number of cards required.
Binary?
“success” = is an ace, “failure” = is not an ace
Independent?
Since the card is being replaced, the probability is not affected.
Number?
There has not been a fix number of trials in advance. You could get an ace the first time, or many times later.
This is not a binomial setting since there was not a fixed number of trials. B I N S

8. Check Your Understanding
Determine whether the given random variables has a binomial distribution. Justify your answer.
Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and shuffle again. Repeat this process 10 times. Let X = the number of aces you observe.
Roll a fair die 100 times. Sometime during the 100 rolls, one corner of the die chips off. Let W = number of 5’s you roll.
1: Binomial 2: not binomial, success is not the same.
Lesson 6.1.1

9. Binomial and Geometric Random Variables
Definition:
The count X of successes in a binomial setting is a binomial random variable. The probability distribution of X is a binomial distribution with parameters n and p, where n is the number of trials of the chance process and p is the probability of a success on any one trial. The possible values of X are the whole numbers from 0 to n.
Binomial and Geometric Random Variables
Note: When checking the Binomial condition, be sure to check the BINS and make sure you’re being asked to count the number of successes in a certain number of trials!

10. Binomial Coefficient

11. Binomial Coefficient

12. Therefore, P(X = 2) = 10(0.25)2(0.75)3 = 0.2637
Binomial Probabilities
Each child of a particular pair of parents has probability 0.25 of having type O blood. Genetics says that children receive genes from each of their parents independently. If these parents have 5 children, the count X of children with type O blood is a binomial random variable with n = 5 trials and probability p = 0.25 of a success on each trial. In this setting, a child with type O blood is a “success” (S) and a child with another blood type is a “failure” (F).
What’s P(X = 2)?
P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 =
However, there are a number of different arrangements in which 2 out of the 5 children have type O blood:
SSFFF
SFSFF
SFFSF
SFFFS
FSSFF
FSFSF
FSFFS
FFSSF
FFSFS
FFFSS
Verify that in each arrangement, P(X = 2) = (0.25)2(0.75)3 =
Therefore, P(X = 2) = 10(0.25)2(0.75)3 =

13. Binomial Coefficient: With TI-83

14. Binomial and Geometric Random Variables
Binomial Probability
The binomial coefficient counts the number of different ways in which k successes can be arranged among n trials. The binomial probability P(X = k) is this count multiplied by the probability of any one specific arrangement of the k successes.
Binomial and Geometric Random Variables
Binomial Probability
the possible values of X are 0, 1, 2, …, n. If k is any one of these values,
Probability of k successes
Probability of n-k failures
Number of arrangements of k successes

15. Lets take a look at those 5 kids and blood type O
Binomial Probability
If X has the binomial distribution with n trials and probability p of success on each trial, the possible values of X are 0,1,2…n. If k is any one of these values,
With our formula in hand, we can now calculate any binomial probability!
Lets take a look at those 5 kids and blood type O

16. binompdf(n,p,k) computes P(X = k)
Each child of a particular pair of parents has probability 0.25 of having type O blood. Genetics says that children receive genes from each of their parents independently. If these parents have 5 children, the count X of children with type O blood is a binomial random variable with n = 5 trials and probability p = 0.25 of a success on each trial. In this setting, a child with type O blood is a “success” (S) and a child with another blood type is a “failure” (F).
What’s P(X = 2)? The chance parents have 2 type O Blood kids
What’s P(X < 2)? The chance parents have at most 2 type O Blood kids
What’s P(X> 2)? The chance Parents have more than 2 type O Blood kids
There are two handy commands on the TI-83/84 for finding binomial probabilities:
binompdf(n,p,k) computes P(X = k)

17. Binomial Probability with TI-83
For example we want to find probability of exactly 2 children with blood type O.
P(X = 2), n = 5, p =0.25, k=2
TI-83:
2nd > VARS >binompdf(> fill in the information> binompdf(5,0.25,2)
TI-89:
In the CATALOG under Flash Apps
What if I wanted to find out P(X>3)???

18. What’s P(X = 2)? The chance parents have 2 type O Blood kids
Binomial Probability On The Calculator
There are two handy commands on the TI-83/84 for finding binomial probabilities:
binompdf(n,p,k) computes P(X = k)
binomcdf(n,p,k) computes P(X ≤ k)
For the parents having n = 5 children, each with probability p = 0.25 of type O blood:
What’s P(X = 2)? The chance parents have 2 type O Blood kids
What’s P(X < 2)? The chance parents have at most 2 type O Blood kids
What’s P(X>2)? The chance Parents have more than 2 type O Blood kids
Should a parent be surprised if they have more than three children with type O Blood?

19. What if I wanted to find out P(X>3)???

20. Check Your Understanding
To introduce her class to binomial distributions, Mrs. Desai gives a 10-item, multiple choice quiz. The catch is, students must simply guess an answer (A through E) for each question. Mrs. Desai uses her computer’s random number generator to produce the answer key, so that each possible answer as an equal chance to be chosen. Patti is one of the students in this class. Let 𝑋= 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑎𝑡𝑡𝑖’𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑔𝑢𝑒𝑠𝑠𝑒𝑠
Show that X is a binomial random variable.
Find 𝑃(𝑋=3), explain what this result means.
To get a passing score on the quiz, a student must guess correctly at least 6 times. Would you be surprised if Patti earned a passing score? Compute an appropriate probability to support your answer.

21. Check Your Understanding
Show that X is a binomial random variable.
Binary? Success: Correct answer Failure: wrong answer
Independent? Each answer does not affect the other.
Number? n=10 problems
Success? p= 1 5 =.20
Find P(X=3), explain what this result means.
Binompdf(10, .2, 3) = .2013, where n=10, p=.2, k=3. There is a 20.13% chance that Patti will answer exactly 3 questions correctly.
To get a passing score on the quiz, a student must guess correctly at least 6 times. Would you be surprised if Patti earned a passing score? Compute an appropriate probability to support your answer.
1 – binomcdf(10, .2, 5) = .0064, where n=10, p=.2, k=5. Since there is only a .64% chance that a student will pass, we would be quite surprised if Patti was able to pass.

22. Mean and Standard Deviation of Binomial Dist.

23. For the parents having 𝑛 = 4 children, each with probability 𝑝 = 0
For the parents having 𝑛 = 4 children, each with probability 𝑝 = 0.25 of type O blood:
Find the mean and standard deviation of X.
Since X is a binomial random variable with parameters 𝑛 = 4 and 𝑝 = .25, we can use the formulas for the mean and standard deviation of a binomial random variable.
If parents who had 4 children were repeatedly observed we’d expect about one fourth of their 4 children to have blood type O, so 1 child to have type O blood on average.
If parents who had 4 children were repeatedly observed, the number of children with type O blood would vary from 1 by .866

24. “10% condition” sampling w/o replacement

25. Objectives Binomial Random Variables and Binomial Distribution
Requirements to be Binomial- B.I.N.S
Binomial Coefficient: by formula and TI-83
Binomial Probability: by formula and TI-83
Mean and Standard Deviation
“10% condition” sampling w/o replacement
Geometric Random Variables and Geometric Distribution
Requirements to be Geometric- B.I.T.S
Geometric Probability: by formula and TI-83
Mean (expected value)

26. Geometric Random Variables
In a binomial setting, the number of trials n is fixed in advance, and the binomial random variable X counts the number of successes. The possible values of X are 0,1,2,….,n. In other situations, the goal is to repeat a chance process until success occurs.

27. Consider this: Roll a pair of dice until you get doubles.
In basketball, attempt a three-point shot until you make one.
Keep placing $1 bet on the number 15 in roulette until you win.
These are all examples of a geometric setting. Although the number of trials isn’t fixed in advance, the trials are independent and the probability of success remains constant.

28. Geometric Random Variables and Geometric Distribution
The number of trials Y that it takes to get a success in geometric setting is a geometric random variable. The probability distribution of Y is a geometric distribution with parameters p, the probability of a success on any trial. The possible values of Y are 1,2,3…

29. Binomial and Geometric Random Variables
Geometric Settings
Definition:
A geometric setting arises when we perform independent trials of the same chance process and record the number of trials until a particular outcome occurs. The four conditions for a geometric setting are
Binomial and Geometric Random Variables
• Binary? The possible outcomes of each trial can be classified as “success” or “failure.” B I
• Independent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial. T
• Trials? The goal is to count the number of trials until the first success occurs. S
• Success? On each trial, the probability p of success must be the same.

30. Requirements to be Geometric- B.I.T.S
A geometric setting arises when we preform independent trials of the same chance process and record the number of trials until a particular outcome occurs.
The four conditions for a geometric setting are:
Binary? The possible outcomes of each trial can be classified as “success” or “failure”
Independent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial
Trials? The goal is the count the number of trials until the first success occurs.
Success? On each trial, the probability p of success must be the same.

31. The Birth Day Game
Your teacher is planning to give you 10 problems for homework. As an alternative, you can agree to play the Birth Day Game. Here’s how it works. A student will be selected at random from your class and asked to guess the day of the week (for instance, Thursday) on which one of your teacher's friends was born. If the student guesses correctly, then the class will have only one homework problem. If the student guesses the wrong day of the week, your teacher will once again select a student from the class at random. The chosen student will try to guess the day of the week on which a different one of your teacher’s friends was born. If this student gets it right, the class will have two homework problems. The game continues until a student correctly guesses the day on which one of your teacher's many friends was born. Your teacher will assign a number of homework problems that is equal to the total number of guesses made by members of your class. Are you ready to play the Birth Day Game?

32. Binomial and Geometric Random Variables
Geometric Settings
Birthday Game Geometric?
Binomial and Geometric Random Variables
• Binary? “success” = correct guess, “failure” = incorrect guess B I
• Independent? student guess has no influence on another guess, since it’s a different person’s birthday. T
• Trials? We are counting the number of trials up and including the first correct guess. S
• Success? The probability of a success 𝑖𝑠 𝑝= 1 7

33. Geometric Probability

34. The Birth Day Game

35. Geometric Probability with TI-83

36. Geometric Probability with TI-83
For example we want to find probability that the class gets exactly 10 homework problems
We want P(Y = 10)….. p =1/7, k=10
TI-83:
2nd > VARS >geometpdf(> fill in the information> geometpdf(1/7, 10)
TI-89:
In the CATALOG under Flash Apps
What if I wanted to find out P(Y<10)???

37. Mean of Geometric Dist.

38. Objectives Binomial Random Variables and Binomial Distribution
Requirements to be Binomial- B.I.N.S
Binomial Coefficient: by formula and TI-83
Binomial Probability: by formula and TI-83
Mean and Standard Deviation
“10% condition” sampling w/o replacement
Geometric Random Variables and Geometric Distribution
Requirements to be Geometric- B.I.T.S
Geometric Probability: by formula and TI-83
Mean (expected value)

39. Finish Chapter 6 Reading Guide
Homework
6.3 homework worksheet
Finish Chapter 6 Reading Guide