1. + Binomial and Geometric Random Variables Textbook Section 6.3

2. + Binomial Settings and Binomial Random Variables What do the following settings have in common? Toss a coin 5 times. Count the number of heads. Spin a roulette wheel 8 times. Count how many times the ball lands in a red slot. Take a random sample of 100 babies. Count the number of females. 1. We have repeated trials of the same chance process & the number of trials is fixed in advance. 2. The trials are all independent. 3. We’re interested in how many times a specific event occurred (we call it “success”) 4. Our chances of getting a “success” are the same for each trial. When ALL these conditions are met, we have a BINOMIAL SETTING.

3. + Binomial Setting details When checking the Binomial conditions, note that there can be more than two possible outcomes per trial. We simply define “success” as one of those outcomes, and all others become “failures” The number of successes we count is a binomial random variable X. The probability distribution of X is called a binomial distribution with parameters n and p. N is the number of trials in the setting P is the probability of success in each trial B(n, p) is the shorthand notation for a binomial setting.

4. + Check your Understanding. Binomial or No?? 1. Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and shuffle again. Repeat the process 10 times. Let X = the number of aces you observe. 2. Choose students at random from your class. Let Y = the number who are over 6 feet tall. 3. Roll a fair die 100 times. Sometime during the 100 rolls, one corner of the die chips off. Let W = the number of 5s you roll.

5. + Binomial Probabilities What if we wanted to find the probability of a couple having two of their five children born with type O blood? The probability of being born with Type O blood for this family is 0.25. Probability of Success = 0.25 and Probability of Failure is 0.75. How many arrangements of 2 “successes” can be made from five children? SFSFF, FSSFF, SSFFF, FFSSF, FSFSF, etc. P(X = 2) = 10(0.25) 2 (0.75) 3 = 0.26367 General Binomial Probability formula:

6. + Using the Calculator for Binomial Probabilities Press 2 nd and VARS buttons for the Distributions menu. Scroll down to find binompdf and binomcdf To find the probability of exactly x number of successes, use binompdf(n, p, x). To find the probability of x or fewer number of successes, use binomcdf(n, p, x) The “c” stands for cumulative and will calculate the probability of X = 0 up to the number desired. P(X = 2) : binompdf (5, 0.25, 2) = 0.26367 P(X ≤ 2) : binomcdf (5, 0.25, 2) = 0.8964 P(X > 4) = 1 − P(X ≤ 3) = 1 – binomcdf (5, 0.25, 3) = 0.0156

7. + Mean and Standard Deviation of Binomial Distributions Binomial Distributions are always skewed right, because the larger values of X are always less likely than smaller values. Mean of a binomial distribution μ X = np Standard Deviation of a binomial distribution = square root of variance. Variance = np(1 – p) Example: On a ten-item multiple choice quiz, students must randomly guess A through E and the teacher uses a random number generator to make the key. Let X = the number of correct answers. Define n & p Find μ X Find standard deviation. What is the probability a student’s number of correct guesses is more than 2 standard deviations above the mean?

8. + Binomial Distributions in Statistical Sampling Example: A flash drive manufacturer inspects an SRS of 10 flash drives from a shipment of 10,000 flash drives. Suppose that 2% of the flash drives are defective. Count the number of bad flash drives in the sample. This is not a strictly binomial setting, because each time a flash drive is selected, the probability changes slightly for the next one. However, figuring the probability of no defective flash drives using a binomial setting = 0.8171 and figuring actual probability is 0.8170 – SO CLOSE!! When the population is much larger than the sample, an SRS can approximate a binomial setting. As a general rule, “much larger” means the sample is no more than 10% of the population!!

9. + Normal Approximation to binomial distributions We already discussed how binomial distributions are skewed, BUT as n gets larger, the distribution gets closer to a Normal Distribution which allows us to use Normal calculations of probability. As a rule of thumb, we will use the Normal approximation when n is so large that np ≥ 10 AND n(1 – p) ≥ 10

10. + Example Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time- consuming.” Suppose that exactly 60% of all respondents would say “Agree.” Let X = the number in the sample who agree. Is this Binomial? Check BINS Check the conditions for using the Normal approximation in this setting. np = (0.6)(2500) = 1500 AND n(1 – p) = (0.4)(2500) = 1000 Use a Normal Distribution to estimate the probability that 1520 or more of the sample agree. μ X = (0.6)(2500) = 1500 and σ X = 24.49 Normalcdf (1520, 10^99, 1500, 24.49) = 0.2071 There is about a 21% chance of getting a sample in which 1520 people or more agree with the statement.

11. + Geometric Random Variables In a binomial setting, the number of trials is fixed in advance. In other situations, the goal is to repeat a chance process until a success occurs. Roll a pair of dice until you get doubles In basketball, attempt a three-point shot until you make one Keep placing a $1 bet on the number 15 in roulette until you win These are all examples of a geometric setting. All other conditions for a binomial setting apply. The number of trials that it takes to get a success in a geometric setting is a geometric random variable.

12. + Geometric Probabilities The general formula for P(X = k) = (1 – p) k – 1 (p) Example: How many times to roll a fair die before you roll a 5? What is the probability it takes 4 rolls? If it took four rolls, you failed three times and succeed on the fourth – P(X = 4) = (5/6) 3 (1/6) To calculate the probability of getting a 5 on or before the fourth roll: P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) or geometcdf (1/6, 4) Mean of a Geometric Distribution (also the expected value or number of trials “expected” before a success occurs): μ X = 1/p