Binomial Probability Distributions PDF

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 outc...

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 pass/fail, 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, or S and failure, or F). 4. The probability of a success remains the same in all trials. Helpful Rule for Independence However: Remember: Two events are independent if one event If your sample size occurring doesn’t effect represents less than 5% the probability of the of the total population, other event. you can treat the This is an issue if you are events as if they are dealing with no independent. replacement.  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) The word success is arbitrary and does not necessarily represent something GOOD. If you are trying to find the probability of deaths from hang-gliding, the ‘success’ probability is that of dying from hang-gliding. Please carefully look at the probability (percentage or rate) provided and whether it matches the probability desired in the question. It might be necessary to determine the complement of the given probability to establish the ‘p’ value of the problem.  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 1 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. 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)  Using Build-In Formula Binomial probability can be calculated with a build-in formula in a graphing calculator. 1. Press “2nd” “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) )  Important Hints ❖ Be sure that x and p both refer to the same category being called a success. ❖ When sampling without replacement, consider events to be independent if n < 0.05N. Other Options Your Saving Grace Your Toolbox Press 2nd VARS (to get 1. Use a TI-83/84 Plus. DISTR), then select the 2. Use the Table A-1. option identified as 3. Use the binomial binompdf. probability table. Complete the entry of binompdf(n, p, x) with specific values for n, p, x, then press ENTER.  Example 2 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) = 5C3 (0.75)3 (0.25)5-3 ≈ 0.2637 Using the Calculator to Find Probability Keyword: “exactly” binompdf(n,p,x) Keywords: “less than or equal to” binomcdf(n,p,x) Keywords: “greater than or equal to” OR “at least” Use the complement: 1- binomcdf(n,p,x) *The “c” in “cdf” stands for “cumulative” Example 3: a) Calculate the probability of tossing a coin 20 times and getting exactly 9 heads. P(exactly 9 heads) = binompdf(20,0.5,9) = 0.160 b) Calculate the probability of tossing a coin 20 times and getting less than 6 heads. less than 6 is equivalent to less than or equal to 5 P (x < 6) = binomcdf(20,0.5,5) = 0.021 c) Calculate the probability of tossing a coin 32 times and getting at least 14 heads. At least 14 is equivalent to 14 or more. Use the complement: 1 – P (x ≤13) 1 – P (x ≤ 13 heads) = 1 – binomcdf(32, 0.5, 13) = 0.811 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 (exactly 1 allergic) = P (1) = binompdf(25,0.01,1) = 0.196 e) Refer to (d), what is the probability that 4 or more of them are allergic to bee stings? 4 or more means at least 4, so use the complement: 1 – P (x ≤ 3) 1 – P (x ≤ 3 allergic) = 1 – binomcdf(25, 0.01, 3) = 0.000107 Example 4: 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 (exactly 5 receive fin. aid) = P (x = 5) =binompdf(9,0.53,5) = 0.257 b) Now using (a), what is the probability that fewer than 3 students in the class receive financial aid? Fewer than 3 is equivalent to less than or equal to 2 P (x ≤ 2) = binomcdf(9,0.53,2) = 0.064 Example 5: Use the following scenario to answer the questions: Consider treating 863 subjects with Lipitor Binomial probability distributions are important (Atorvastatin) becauseand theyrecording allow us to whether there is a “yes” deal with circumstances in responsewhich whenthethey outcomes belong are asked iftothey TWOexperienced categories, such a as pass/fall, acceptable/defective, etc. 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 6: 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 Example 7: 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.  Methods for Finding Probabilities We will now discuss the methods for finding the probabilities corresponding to the random variable x in a binomial distribution. Example 8: 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 Table Table results can be used to find binomial probabilities. Table Example 9: 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. x P(x) a) Find the probability that at least 1 of the 0 0.050328 5 donors has Group O blood. 1 0.205889 P(X ≥ 1) = 1 – P(X = 0) = 1 – 0.050328 = 0.949672 2 0.336909 P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) 3 0.275653 + P(X = 4) + P(X = 5) = 0.949671 4 0.112767 b) If at least 1 Group O donor is needed, 5 0.018453 is it reasonable to expect that at least 1 will be obtained? Yes, since P(X ≥ 1) = 0.949671. Example 10: 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 11: 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 McDonald’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 McDonald’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: ❖ The definition of the binomial probability distribution. ❖ Notation. ❖ Important hints. ❖ Three computational methods. ❖ 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.