# Binomial & Geometric Random Variables

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

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)

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)

Binomial Random Variables and Binomial Distribution
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.

Requirements to be Binomial- B.I.N.S
A binomial setting arises when we preform 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: 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 Number? The number of trials n of the chance process must be fixed in advance. Success? On each trial, the probability p of success must be the same.

Examples: Here are 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. Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y= the number of Aces you observe. 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.

Solutions to 3 Scenarios:
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

Solutions to 3 Scenarios:
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. 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. 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.

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. Chose three students at random from your class, without replacement. let Y = the number who are over 6 feet tall. Flip a coin. If heads, roll a 6-sided die. If its tails, roll an 8-sided die. Repeat this process 5 times. Let W= the number of 5’s you roll.

Binomial Coefficient

Binomial Coefficient: With TI-83

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!

Inheriting Blood Type Each child of a particular pair of parents has probability 0.25 of having type O blood. Suppose the parents have 5 children. Find the probability that exactly 3 of the children have type O blood. Should the parents be surprised if more than 3 of their children have type O blood? Justify your answer.

Binomial Probability with TI-83

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

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

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.

Show that X is a binomial random variable. Binary? Independent? Number? Success? Find P(X=3), explain what this result means. Binompdf(10, .2, 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, since there is only a .64% chance that a student will pass, we would be quite surprised if Patti was able to pass.

Mean and Standard Deviation of Binomial Dist.

“10% condition” sampling w/o replacement

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.

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.

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…

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.

Birthday Game Geometric?
Binary? “success” = correct guess, “failure” = incorrect guess Independent? – student guess has no influence on another guess, since it’s a different person’s birthday. Trials? We are counting the number of trials up and including the first correct guess. Success? The probability of a success is p=1/7 This is a geometric setting

Geometric Probability

The Birth Day Game

Geometric Probability with TI-83

Geometric Probability with TI-83
For example we want to find probability that the class gets exactly 10 homework problems 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)???

Mean of Geometric Dist.

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)

Test Results! Grade: Amount: Marginal % ……A......……....6.……….31%
…….B…………...5……...26% 73% Passed …….C…………..3..……...16% …….D…………..3.……...16% …….F… ………..11% 27% Failed Mean: 79% Max: 100% Min: 50% No Outliers

Tracking AP Stats 2014-2015 (WHS)
Ch. 1 Test Ch. 2 Test Ch. 3 Test Ch. 4 Test Ch. 5 Test A 5 A 5 A3 A1 A6 B5 B 6 B5 B9 B5 C6 C 4 C6 C3 C3 D2 D 1 D2 D5 D3 F1 F 2 F2 F1 F2

Homework Worksheet