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Understanding Binomial Probability Models for Real-World Scenarios

This text delves into the concept of binomial random variables and probability distributions, offering explanations, examples, calculations, and applications in various scenarios such as sports, taste tests, and color blindness studies. Learn how to calculate probabilities for different outcomes using binomial formulas and enhance your understanding through practical examples.

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Understanding Binomial Probability Models for Real-World Scenarios

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  1. A Useful Probability Model:Binomial Random Variables Binomial Probability Distributions

  2. Warmup • Challenging job interview questions What is the probability that an integer between 50,000 and 59,999 has exactly two 6’s?

  3. Binomial Random Variables • Through 2/16/2019 NC State’s free-throw percentage is 70.4% (173rd out 351 in Div. 1). • If in the 2/20/2019 game with BC, NCSU shoots 11 free-throws, what is the probability that: • NCSU makes exactly 8 free-throws? • NCSU makes at most 8 free throws? • NCSU makes at least 8 free-throws?

  4. Fans Flip Out Over NFL's Rule on Coin Toss and Overtime. Is It Fair? Warmup As luck would have it, the Patriots called heads, then beat the Chiefs in overtime.Wall Street Journal, Feb. 1, 2019

  5. “2-outcome” situations are very common • Heads/tails • Democrat/Republican • Male/Female • Win/Loss • Success/Failure • Defective/Nondefective

  6. Probability Model for this Common Situation • Common characteristics • repeated “trials” • 2 outcomes on each trial • Leads to Binomial Experiment

  7. Binomial Experiments • n identical trials • n specified in advance • 2 outcomes on each trial • usually referred to as “success” and “failure” • p “success” probability; q=1-p “failure” probability; remain constant from trial to trial • trials are independent

  8. Classic binomial experiment: tossing acoin a pre-specified number of times • Toss a coin 10 times • Result of each toss: head or tail (designate one of the outcomes as a success, the other as a failure; makes no difference) • P(head) and P(tail) are the same on each toss • trials are independent • if you obtained 9 heads in a row, P(head) and P(tail) on toss 10 are same as P(head) and P(tail) on any other toss (not due for a tail on toss 10)

  9. Binomial Random Variable • The binomial random variable X is the number of “successes” in the n trials • Notation: X has a B(n, p) distribution, where n is the number of trials and p is the success probability on each trial.

  10. Binomial Probability Distribution

  11. Rationale for the Binomial Probability Formula n! P(x) = •px•qn-x (n –x )!x! Number of outcomes with exactly x successes among n trials

  12. Binomial Probability Formula n! P(x) = •px•qn-x (n –x )!x! Probability of x successes among n trials for any one particular order Number of outcomes with exactly x successes among n trials

  13. The sum of all the areas is 1 p(5)=.246 is the area of the rectangle above 5 Graph of p(x); x binomial n=10 p=.5; p(0)+p(1)+ … +p(10)=1 Think of p(x) as the area of rectangle above x

  14. Binomial DistributionExample: Pepsi vs Coke • In a taste test of Pepsi vs Coke, suppose 25% of tasters can correctly identify which cola they are drinking. • If 12 tasters participate in a test by drinking from 2 cups in which 1 cup contains Coke and the other cup contains Pepsi, what is the probability that exactly 5 tasters will correctly identify the colas?

  15. Binomial Distribution Example • ShanilleO’Keal is a WNBA player who makes 25% of her 3-point attempts. • Assume the outcomes of 3-point shots are independent. • If Shanille attempts 7 3-point shots in a game, what is the expected number of successful 3-point attempts? • Shanille’s cousin Shaquille O’Neal makes 10% of his 3-point attempts. If they each take 12 3-point shots, who has the smaller probability of making 4 or fewer 3-point shots? Shanille has the smaller probability.

  16. 9, 10, 11, … , 20 Using binomial tables; n=20, p=.3 • P(x  5) = .416 • P(x > 8) = 1- P(x  8)= 1- .887=.113 • P(x < 9) = ? • P(x  10) = ? • P(3  x  7)=P(x  7) - P(x  2) .772 - .035 = .737 8, 7, 6, … , 0 =P(x 8) 1- P(x  9) = 1- .952

  17. Binomial n = 20, p = .3 (cont.) • P(2 < x  9) = P(x  9) - P(x  2) = .952 - .035 = .917 • P(x = 8) = P(x  8) - P(x  7) = .887 - .772 = .115

  18. Color blindness The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population. We can model this situation with a B(n = 25, p = 0.08) distribution. • What is the probability that five individuals or fewer in the sample are color blind? Use Excel’s “=BINOMDIST(number_s,trials,probability_s,cumulative)” P(x≤ 5) = BINOMDIST(5, 25, .08, 1) = 0.9877 • What is the probability that more than five will be color blind? P(x> 5) = 1 P(x≤ 5) =1  0.9877 = 0.0123 • What is the probability that exactly five will be color blind? P(x= 5) = BINOMDIST(5, 25, .08, 0) = 0.0329

  19. B(n = 25, p = 0.08) Probability distribution and histogram for the number of color blind individuals among 25 Caucasian males.

  20. What are the expected value and standard deviation of the count X of color blind individuals in the SRS of 25 Caucasian American males? E(X) = np = 25*0.08 = 2 SD(X) = √np(1 p) = √(25*0.08*0.92) = 1.36 What if we take an SRS of size 10? Of size 75? E(X) = 10*0.08 = 0.8 E(X) = 75*0.08 = 6 SD(X) = √(10*0.08*0.92) = 0.86 SD(X) = (75*0.08*0.92)=2.35 p = .08 n = 10 p = .08 n = 75

  21. Recall Free-throw question • Through 2/16/2019 NC State’s free-throw percentage is 70.4% (173rd out of 351 in Div. 1). • If in the 2/20/2019 game with BC, NCSU shoots 11 free-throws, what is the probability that: • NCSU makes exactly 8 free-throws? • NCSU makes at most 8 free throws? • NCSU makes at least 8 free-throws? • n=11; Success=make free throw; p=.704 X=# of made free-throws. p(8)= 11C8 (.704)8(.296)3 =.258… • P(x ≤ 8)=.676… • P(x ≥ 8)=1-P(x≤7) =1-.4187 = .5813

  22. Recall Warmup Question • Challenging job interview questions What is the probability that an integer between 50,000 and 59,999 has exactly two 6’s? Ten equally likely choices for each of the four digits 5?, ? ? ? 0, …, 9 We want the probability of 2 successes in 4 trials Success: select a 6 n=4 p(S)=0.1

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