Section 5 3 Binomial Probability Distributions Learning Targets

Section 5. 3 Binomial Probability Distributions

Learning Targets This section presents a basic definition of a binomial distribution along with notation, and methods for finding probability values. Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two relevant categories such as acceptable/defective or survived/died.

Binomial Probability Distribution A binomial probability distribution results from a procedure that meets all the following requirements: 1. The procedure has a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials. ) *If a sample is less than 5% of the population, it may be treated as independent. 3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure). 4. The probability of a success remains the same in all trials.

Notation for Binomial Probability Distributions S and F (success and failure) denote the two possible categories of all outcomes; p and q will denote the probabilities of S and F, respectively, so P(S) = p (p = probability of success) P(F) = 1 – p = q (q = probability of failure)

Notation (continued) n denotes the fixed number of trials. x denotes a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive. p denotes the probability of success in one of the n trials. q denotes the probability of failure in one of the n trials. P(x) denotes the probability of getting exactly x successes among the n trials.

Example �Consider an experiment in which 5 offspring peas are generated from 2 parents such that there is a 75% chance of an individual pea having a green pod and a 25% chance of it having a yellow pod. �Is this a Binomial Probability Distribution? �Yes.

Example �Consider the earlier experiment in which 5 offspring peas are generated from 2 parents such that there is a 75% chance of an individual pea having a green pod and a 25% chance of it having a yellow pod. �What’s the probability of getting exactly 3 offspring peas with green pod’s? P(3) = 5 C 3 • (0. 75)3 • (0. 25)5 -3 0. 2637

Example 1: Use the following scenario to answer the Binomial probability distributions are important because questions: Consider treating 863 subjects with Lipitor they allow us to deal with circumstances in which the (Atorvastatin) and recording whethere is a “yes” outcomes belong to TWO categories, such as pass/fall, acceptable/defective, etc. response when they are asked if they experienced a headache (based on data from Pfizer, Inc. ). Suppose we want to find the probability that 571 said “yes”. a) Determine whether or not the given procedure results in a binomial distribution. If it does not binomial, identify at least one requirement that is not satisfied. 1. Fixed number of trials (863), 2. Subject are independent from each other, 3. Each trial only answer Y/N, 4. The probability of Y/N remains ½ through each trial. b) If it is binomial, identify the values of n, x, p, and q. n = 863, x = 571. p = ½, q = ½

Example 2: Use the following scenario to answer the questions: Consider treating 152 couples with YSORT gender selection method developed by the Genetics & IVF Institute and recording the ages of the parents. Suppose we want to find the probability of choosing a parent the age of 30. a) Determine whether or not the given procedure results in a binomial distribution. If it does not binomial, identify at least one requirement that is not satisfied. 1. Fixed number of trials (152), 2. Subject are independent from each other, 3. Each trial has more than 2 age levels MUST MEET ALL REQUIREMENTS!!! b) If it is binomial, identify the values of n, x, p, and q. N/A

Helpful Rule for Independence However: Remember: � Two events are independent if one event occurring doesn’t effect the probability of the other event. � This is an issue if you are dealing with no replacement. � If your sample size represents less than 5% of the total population, you can treat the events as if they are independent.

Example 3: Twenty different Senators are randomly selected from the 100 Senators in current Congress, and each was asked whether he or she is in favor of abolishing state taxes. Suppose we want to find the probability that 15 Senators are in favor. a) Determine whether or not the given procedure results in a binomial distribution. If it does not binomial, identify at least one requirement that is not satisfied. b) If it is binomial, identify the values of n, x, p, and q. Answer: Not binomial(because trials are NOT Independent). The Senators are chosen without replacement, so the events are not independent. Can we use the 5% guideline? 20/100 = 0. 20 > 0. 05 Our sample is more than 5%, so the 5% guideline cannot be applied.

Important Hints v Be sure that x and p both refer to the same category being called a success. v When sampling without replacement, consider events to be independent if n < 0. 05 N.

Methods for Finding Probabilities We will now discuss four methods for finding the probabilities corresponding to the random variable x in a binomial distribution.

A formula for Calculating Probability The Formula � The Components � n: the number of trials � x: the number of successful trials � p: the probability of success � q: the probability of failure ( q = 1 – p)

Binomial Probability Formula Number of outcomes with exactly x successes among n trials Calculator: 1. 2 nd + “VARS” keys 2. Select “ binompdf( ” 3. Enter trial = n, p, x value = x The probability of x successes among n trials for any one particular order The remaining two factors of the formula will compute the probability of any one arrangement of successes and failures. This probability will be the same no matter what the arrangement is. The three factors multiplied together give the correct probability of ‘x’ successes.

Other Options Your Saving Grace � Press 2 nd VARS (to get DISTR), then select the option identified as binompdf. � Complete the entry of binompdf(n, p, x) with specific values for n, p, x, then press ENTER. Your Toolbox 1. 2. 3. Use a TI-83/84 Plus. Use the Table A-1. Use the binomial probability table.

Using Build-In Formula � Binomial probability can be calculated with a build-in formula in a graphing calculator. Press “ 2 nd” “VARS” 2. Select “binompdf (” 3. Put n in “trials: ”, p in “p: ”, x in “x value: ”. � (In TI-83, you may put n, p, x in such order into binompdf ( n, p, x) ) 1.

Example 4: Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. a) n = 12, x = 10, p = 3/4 P(X = 10) = binnompdf(12, ¾, 10) = 0. 2323 b) n = 20, x = 4, p = 0. 15 P(X = 4) = binnompdf(20, 0. 15, 4) = 0. 1821

Method 2: Using Minitab Table Technology Minitab Table results can be used to find binomial probabilities. Minitab

Example 5: Refer to the Minitab display. (When blood donors were randomly selected, 45% of them had blood that is Group O (based on data from the Greater New York Blood Program). ) The display shows the probabilities obtained by entering the values of n = 5 and p = 0. 45. a) Find the probability that at least 1 of the 5 donors has Group O blood. P(X 1) = 1 – P(X = 0) = 1 – 0. 050328 = 0. 949672 P(X 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0. 949671 b) If at least 1 Group O donor is needed, is it reasonable to expect that at least 1 will be obtained? x P(x) 0 0. 050328 1 0. 205889 2 0. 336909 3 0. 275653 4 0. 112767 5 0. 018453

Method 3: Using TI 83/84 Technology We will be doing binomial probabilities using the calculator: *Keyword “exact” …find probability with binompdf(number of trials, probability of success, number of successes) *Keywords “less than” or “less than or equal to” or “at most…” …find probability with binomcdf(number of trials, probability of success, up to the number of successes) *Keywords “at least” or “greater than or equal to” or “no fewer than…” … find probability with 1 – binomcdf with 1 less number for success (What you are doing here is 1 minus what you DO NOT WANT and that will give you the probability that you do want. )

Example 6: Use the TI 83/84 calculator to calculate the following probabilities. a) Calculate the probability of tossing a coin 20 times and getting exactly 9 heads. P(X = 4) = binnompdf(20, 0. 5, 9) = 0. 1602 b) Calculate the probability of tossing a coin 20 times and getting less than 6 heads. P(X < 6) = P(X 5) = binnomcdf(20, 0. 5, 5) = 0. 0207 c) Calculate the probability of tossing a coin 32 times and getting at least 14 heads. P(X 14) = 1 – P(X < 14) = 1 – binnomcdf(32, 0. 5, 13) = 0. 8115 d) About 1% of people are allergic to bee stings. What is the probability that exactly 1 person in a class of 25 is allergic to bee stings? P(X =1) = binnompdf(25, 0. 01, 1) = 0. 0041 e) Refer to (d), what is the probability that 4 or more of them are allergic to bee stings? P(X 4) = 1 – P(X < 4) = 1 – binnomcdf(25, 0. 01, 3) = 0. 0001

Example 7: The brand name of Mrs. Fields has a 90% recognition rate. If Mrs. Fields herself wants to verify that rate by beginning with a small sample of 10 randomly selected consumers, find the probability that exactly 9 of the 10 consumers recognize her brand name. Also find the probability that at least 7 consumers recognize her brand name. P(X = 9) = binnompdf(10, 0. 9, 9) = 0. 3874 P(X 7) = 1 – P(X < 7) = 1 – binnomcdf(10, 0. 9, 6) = 0. 9872

Your Turn �Mc. Donald’s has a brand name recognition rate of 95%. Assuming that we randomly select 5 people, what is the probability that at least 3 of the 5 have heard of Mc. Donald’s? binompdf (5, 0. 95, 3) + binompdf (5, 0. 95, 4) + binompdf (5, 0. 95, 5) 0. 9988 P(X 3) = 1 – P(X < 3) = 1 – binnomcdf(5, 0. 95, 2) = 0. 9988

Your Turn: a) At a college, 53% of students receive financial aid. In a random group of 9 students, what is the probability that exactly 5 of them receive financial aid? P(X = 5) = binnompdf(9, 0. 53, 5) = 0. 2571 b) Now using (a), what is the probability that fewer than 3 students in the class receive financial aid? P(X < 3) = P(X 2) = binnomcdf(9, 0. 53, 2) = 0. 0637 c) Now using (a), what is the probability that at least 6 students in the class receive financial aid? P(X 6) = 1 – P(X < 6) = 1 – binnomcdf(9, 0. 53, 5) = 0. 3164

Recap In this section we have discussed: v The definition of the binomial probability distribution. v Notation. v Important hints. v Three computational methods. v Rationale for the formula.

Homework �Pg. 225: 5 -8, 15, 16, 21 -22, 31, 36 �Helpful hint: � The probability of “at least” is the same as 1 minus the probability of LESS than that number.