Chapter 17

1. Chapter 17 Probability Models The Binomial Distribution

2. Determine the type of distribution and solve: You roll a fair six-sided die: What is the probability that you will get the first 6 on the 9th roll? This is a geometric distribution (since we wish to determine the first success on the 9th roll). It satisfies all of the conditions, so we will use geometpdf(1/6, 9) ≈ .0388 What is the probability that you will get the first 6 within 9 rolls? This is still a geometric distribution (since we wish to determine the first success in any of the first 9 roll). It satisfies all of the conditions, so we will use geometcdf(1/6, 9) ≈ .8062 The die is rolled 9 times. What is the probability that you will get 6 exactly 4 times? This is a binomial distribution (since it has a fixed number of trials). It satisfies all of the conditions, so we will use binompdf(9, 1/6, 4) ≈ .0391 Practice Problems

3. Determine the type of distribution and solve: You pick a card from a standard 52-card deck and replace the card. You do this 5 times. What is the probability that you will get exactly 3 face cards? (Aces are not face cards) This is a binomial distribution (since it has a fixed number of trials). It satisfies all of the conditions, so we use binompdf(5, 12/52, 3) ≈ .073 If we pick 5 cards from a standard 52-card deck, we expect to see exactly 3 face cards about 7.3% of the time. You pick a card from a standard 52-card deck and replace the card. How many cards will you pick before you get a face card? This is a geometric distribution (since we are interested in getting our first face card). It satisfies all of the conditions, so we use 1/p = 1/(12/52) ≈ 4.333 We expect to see the first face card in about 4.333 picks. Practice Problems

4. Practice Problems • Determine the type of distribution and solve: • You pick 5 card from a standard 52-card deck without replacement. What is the probability that you will get exactly 3 face cards? • This problem is not a Bernoulli Trial since it violates two of our rules: the probabilities are not the same for each card and each card is not independent. So, we need to solve this using another method. • We want to determine the probability of getting 3 face cards and two non-face cards and multiply it by the number of ways of getting 3 face cards out of five:

5. Determine the type of distribution and solve: A new sales gimmick has 30% of the M&M’s covered with speckles. These “groovy” candies are mixed randomly with the normal M&M’s as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for the speckles. What’s the probability that we’ll find two speckled ones if you grab five candies? Since we are looking at finding 2 speckled candies out of five, this is a binomial distribution. What’s the first thing you should do before you try to answer this question? Practice Problems

6. Check the conditions: Two-Outcome Assumption There are two outcomes: success = specklesandfailure = ordinary Consistent Probability Assumption The probability of success, based on the information from the candy company, is 30%. Independence Assumption We want to assume that the trials are independent – there is no reason to believe that finding a speckled candy reveals anything about whether the next one out of the bag will have speckles. However, we are actually drawing M&M’s without replacement from a finite population, so the Independence Condition is actually violated; the probability that the next one is speckled depends on how many speckled ones we have removed from the pool so far. So what condition will allow me to continue? Practice Problems

7. Practice Problems • Recall that the 10% Condition states that if independence is violated, it is acceptable to proceed as long as the sample is smaller than 10% of the population. • The Independence Assumption is violated whenever we sample without replacement, but is overridden by the 10% Condition (as long as we don’t drain off more than 10% of the population, the probabilities don’t change enough to matter). • Back to the M&M problem: • Obviously our sample (the small bag we bought) is such a tiny subset of the vast number of M&Ms in the world that it doesn’t really affect the probabilities much, and that’s the essence of the 10% Condition. • So, we can apply the binomial distribution: P(X = 2) ≈ 0.3087 In the TI-84: binompdf(5, .3, 2)

8. Practice Problems • The owner of a small convenience store is trying to decide whether to discontinue selling magazines. He suspects that only 5% of the customers buy a magazine and thinks that he might be able to use the display space to sell something more profitable. Before making a final decision he decides that for one day he’ll keep track of the number of customers and whether or not they buy a magazine. • What is the probability that exactly 2 of the first 10 customers buy magazines? Assume all the assumptions are satisfied. What distribution should you use? • Since we know that we are looking at 10 customers, this is a binomial distribution. P(X = 2) ≈ 0.0746 • In the TI-84: binompdf(10, .05, 2)

9. Practice Problems • Using the same information as before, what is the probability that at least 5 of his first 50 customers buy magazines? • P(X ≥ 5) = 1 – binomcdf(50, .05, 4) ≈ 0.1036 • Using the same information as before, what is the probability that 4 to 7 of his first 50 customers buy magazines? • Since we want 4 to 7, we want X = 4, 5, 6, or 7 • SoP(4 ≤ X ≤ 7) ≈ 0.2364 =binomcdf(50, .05, 7) – binomcdf(50, .05, 3) ≈ 0.2364

10. Practice Problems • The store had 2752 customers that day. Assuming this day was typical for his store, what would be the mean and standard deviation of the number of customers who buy magazines each day? • For simplicities sake, we often use q to represent (1 – p) so

11. Practice Problems • The store had 2752 customers that day. Using the normal model, what is the approximate probability that less than 125 people will buy magazines? What condition do you need to check to solve this problem? • The condition that allows us to use the normal model is… • Success/Failure Condition: np ≥ 10 and nq ≥ 10 • 2752(.05) = 137.6 > 10 and 2752(.95)=2614.4 > 10, since we expect to see more than 10 successes and failures, it’s alright to use the normal model to approximate the probabilities.

12. Practice Problems The store had 2752 customers that day. Using the normal model, what is the approximate probability that less than 125 people will buy magazines? We know: • If we use the z-score, we get: • Use the normal model to calculate the probability. P(X < 50) = normalcdf (-E99, -1.102) ≈ 0.1352. • So, there is a 13.52% chance that less than 125 people will buy a magazine on any given day. • Or we can simply use normalcdf(-E99, 125, 137.6, 11.4333)

13. What Can Go Wrong? • Be sure you have Bernoulli trials. • You need two outcomes per trial, a constant probability of success, and independence. • Remember that the 10% Condition provides a reasonable substitute for independence. • Don’t confuse Geometric and Binomial models. • Don’t use the Normal approximation with small n. • You need at least 10 successes and 10 failures to use the Normal approximation.

14. What have we learned? • Geometric model • When we’re interested in the number of Bernoulli trials until the next success. • Binomial model • When we’re interested in the number of successes in a certain number of Bernoulli trials. • Normal model • To approximate a Binomial model when we expect at least 10 successes and 10 failures. Note: This gives only a rough approximation.

15. What have we learned? geometpdf(p, x) = probability the first success is on the xth trial geometcdf(p, x) = probability the first success is within the 1st to the xth trial binompdf(n, p, x) = probability of x successes in n trials. binomcdf(n, p, x) = the sum of the probabilities of the first x success in n trials.

16. Assignment