1. More on Binomial Distributions

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1 Math 25-Introductory Statistics Lecture 9/27/06. More on Binomial Distributions When we toss a coin four times, and we compute the probability of getting three heads, for example, we need to know how many ways that can happen. One way is to list out all the different ways, or as we saw in the last lecture, we can use Pascal s triangle. This pattern is the same for all binomial distributions. The specific probabilities are a little different, and we ll work through that today. 2. Rolling a die four times We know how tossing a coin four times and counting up heads works. Now let s look at rolling a die four times and counting up s. We re interested in s and things other than s. This makes this a binomial distribution, since there are two possibilities on each trial. I ll use for the s (of course) and for all the other numbers. How many ways can we get three s? Let s list them out (),,,. There are four ways. Compare this to the ways we can get three heads in four coin tosses. (2) HHHT,HHTH,HTHH,THHH. They re basically the same, and the number of ways is exactly the same. That s not really a surprise. So with rolling a die and counting up s, we can figure out the number of ways using Pascal s triangle, since it s the same as what we did last time. We know this much of the table. (3) Number of s Number of ways Probability 4

2 2 Let s look at computing the probability of exactly one head in four tosses, and then see how getting exactly one is different. (4) P (HTTT or THTT or TTHT or TTTH) = = =4 6 The numbers come out different, but the patterns are similar. Note that the probability of getting a is 6, and the probability of not getting a is 5 6,so (5) P ( or or or ) = = =4 296 The important thing to notice here is that in both cases, all four of the individual probabilities, 25 6 and 296, were the same. This allows us to multiply by 4, the number of ways. As a result, we only have to look at one of the ways in each row to find the appropriate probabilities. Let s take the row for two s. One way to get two s is. The probability for this is (6) P ( )= = The other five ways must have the same probability, so (7) P (two s) = = We can fill out the table with the other probabilities. (8) Number of s Number of ways Probability = = = = = 296

3 3 All the probabilities in a probability distribution should add up to, since we ve covered every possibility. Let s check this one. (9) = =. 3. Quiz, Part I of II Make a probability distribution table for the experiment of rolling a die three times, and counting the number of s. Find the following probabilities. Use the denominator 26 for all fractions.. P (zero s). 2. P (one ). 3. P (two s). 4. P (three s). 4. Guessing on a multiple choice test Any experiment that consists of repeated trials with two outcomes on each trial is a binomial experiment. Let s say you re taking a multiple choice test, and you just guess on each question. This would be a binomial experiment, and we can compute the probabilities. Suppose there are four questions, and each question has four answers to choose from. Each question is a trial, and the probability of getting a right answer by just guessing must be one out of four, or 4. The probability of getting a wrong answer is. The number of ways for the different outcomes, any particular number of correct answers, comes from Pascal s triangle and is the same for any binomial experiment with four trials. The probability distribution table must look like the following. (0) Number correct Number of ways Probability 4 For one correct, we can find the probability as in the earlier examples. There are four ways to get one correct, and one of them is wcww. That is, wrong on the first question, correct

4 4 on the second, and wrong on the third and fourth questions. The probability for this one possibility is () P (wcww)= = The other three ways to get one correct must have this same probability, so we can complete the second row of the table as follows. (2) Number correct Number of ways Probability = Quiz, Part II of II Complete the rest of the table for the multiple choice test. Find the following probabilities. Use the denominator 256 for all fractions. 5. P (0 correct). 6. P ( correct). 7. P (2 correct). 8. P (3 correct). 9. P (4 correct). 0. To make the points come out even, answer Yes for Question Homework For problems -4, the binomial experiment is my brother Andy shooting freethrows. He is a 60% freethrow shooter. We ll say that that means that the probability that Andy will make a particular freethrow is 60% or 3 5. Andy s going to try three freethrows. The probability of a success on each trial is 3 5. Make the probability distribution table, and find the following probabilities. Write your answers as fractions over 25.

5 5. P (0 freethrows). 2. P ( freethrows). 3. P (2 freethrows). 4. P (3 freethrows). Now, Andy s going to try five freethrows. Make the probability distribution table, and find the following probabilities. Write your answers as fractions over 325 (with no commas). 5. P (0 freethrows). 6. P ( freethrows). 7. P (2 freethrows). 8. P (3 freethrows). 9. P (4 freethrows). 0. P (5 freethrows).

= = 0.1%. On the other hand, if there are three winning tickets, then the probability of winning one of these winning tickets must be 3 (1)

= = 0.1%. On the other hand, if there are three winning tickets, then the probability of winning one of these winning tickets must be 3 (1) MA 5 Lecture - Binomial Probabilities Wednesday, April 25, 202. Objectives: Introduce combinations and Pascal s triangle. The Fibonacci sequence had a number pattern that we could analyze in different