= = 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)

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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 ways. We re going to start looking at another number pattern that has explicit and relational forms. Let s start this off by discussing probability briefly. Tossing a coin. If we toss a coin, there are two possible outcomes: a head or a tail. We generally accept the fact that the probability of getting a head is 2. One way of interpreting this probability is that if we repeat the experiment of tossing a coin a number of times, then we would expect heads to come up half of the time. In practice, if we were to toss a thousand coins, we generally see that about 500 of the coins would come up heads. Results like 85 heads or 507 heads would not be surprising. Rolling a die. A die is a cube with six faces numbered one through six. The symmetry of the die suggests to us that one number is no more likely to come up than any other. After a large number of rolls, therefore, we would expect to see s come up about a sixth of the time, 2 s about a sixth of the time, etc. In other words, the probability of a must be 6. Basic Principle. A basic principle of probability is that if we are faced with n equally likely outcomes, then the probability of any one of these outcomes must be n. Lottery. If there are one thousand tickets sold in a lottery, and one of these is a winning ticket, then the probability of buying a winning ticket must be,000 = 0.00 = 0.%. On the other hand, if there are three winning tickets, then the probability of winning one of these winning tickets must be (),000. Basic Principle 2. If you have a combination of equally likely outcomes, then the probabilities add. In other words, if there are n equally likely outcomes, and you re considering a collection of m of these, then the probability that one of these m outcomes will occur is m n. How many heads? Suppose I want to find the probability of getting three heads out of ten coin tosses. I need to find a collection of equally likely outcomes, and then count the number of ways that correspond to three heads. Let s start with a simpler case, and then work up from there. Two tosses. If I toss a coin twice, there are four possible outcomes. Notation st toss 2nd toss TT Tail Tail HT Head Tail TH Tail Head HH Head Head If you think about this, there is no reason to believe that one of these four outcomes is more or less likely than any of the others. We have, therefore, four equally likely outcomes. This means that the probability of getting any one of these outcomes is. The probability of getting a head first and a tail second, for example, must be. We re interested in describing the outcomes differently: by the number of heads. The possibilities are zero heads, one head, and two heads. It is important to note that these outcomes are not equally likely, but we do know

2 MA 5 Lecture - Binomial Probabilities 2 zero heads TT one head HT, TH 2 two heads HH. Consider the case where we have three coin tosses. Fill out the following tables. Notation st toss 2nd toss rd toss TTT Tail Tail Tail HT T Head Tail Tail THT. zero heads TTT 8 one head HTT, THT, TTH 8 two heads three heads The pattern Let me do the next case, and the pattern should start to come more clear. With four tosses, the table looks like zero heads TTTT 6 one head HTTT, THTT, TTHT, TTTH 6 two heads HHTT, HTHT, HTTH, THHT, THTH, TTHH 6 6 three heads HHHT, HHTH, HTHH, THHH 6 four heads HHHH 6 One part of the pattern is that the total number of ways is always a power of two. In the case of four tosses, there are two possibilities for the first toss, two for the second, two for the third, and two for the fourth. This makes (2) = 2 = 6 ways.

3 MA 5 Lecture - Binomial Probabilities Another way to think of this is that there are 2 ways in the three-toss case, and the fourth toss can be H or T for each of these ways. Therefore, there should be twice as many ways as with three tosses, or 2 2 = 2 ways. We also need to figure out the pattern for the numbers of ways to get one head, two heads, etc. Let s start off the pattern with the most trivial cases. zero tosses. If we don t toss a coin, there is only one outcome: one way to get zero heads. one toss. If we toss the coin once, there is one way to get zero heads (T), and one way to get one head (H). We ve already done two tosses, three tosses, and four tosses. These numbers are collected in the following table. zero tosses: one toss: two tosses: 2 three tosses: four tosses: 6 Do you see the pattern? We have a roof of s, and every number is the sum of the two immediately above it. I can extend the table pretty easily now. Here are the next three rows. zero tosses: one toss: two tosses: 2 three tosses: four tosses: 6 five tosses: six tosses: seven tosses: This table is called Pascal s triangle. 2. Extend Pascal s triangle three more rows. The probabilities OK. Let s look at the seven toss row. It goes:, 7, 2, 5,5,2,7,. These are the number of ways to get: zero heads, one head, two heads, three heads, four heads, five heads, six heads, and seven heads. For example, there must be 5 ways to get three heads. There are 2 7 = 28 total possible ways (and so the row should also add up to 28). The probability of getting three heads in seven tosses would be = 27.%. 28. What is the probability of getting three heads out of ten tosses? Practice Problem answers on last page.

4 MA 5 Lecture - Binomial Probabilities Quiz Worksheet You ll want to have the first eleven rows Pascal s triangle sitting in front of you (up to the row that starts, 0,...). Use Pascal s triangle to answer the following questions.. How many ways are there to toss a coin seven times? 2. How many ways are there to get three heads out of seven tosses?. What is the probability of getting three heads out of seven tosses? Write your answer as a fraction, like 50/28 You don t have to reduce your fractions, but you can, if you wish.. How many ways are there to get five heads out of seven tosses? 5. What is the probability of getting five heads out of seven tosses? 6. How many ways are there to toss a coin ten times? 7. How many ways are there to get three heads out of ten tosses? 8. What is the probability of getting three heads out of ten tosses? 9. What is the probability of getting seven heads out of ten tosses? 0. What is the probability of getting ten heads out of ten tosses?

5 MA 5 Lecture - Binomial Probabilities 5 ) Your first column should be TTT, HTT, THT, TTH, HHT, HTH, THH, HHH. You can have a different ordering, but the pattern you use should make sense to you. two heads: HHT, HTH, THH 8. three heads: HHH 8. 2) eight tosses:, 8, 28, 56, 70, 56, 28, 8, nine tosses:, 9, 6, 8, 26, 26, 8, 6, 9, ten tosses:, 0, 5, 20, 20, 252, 20, 20, 5, 0, ) From Pascal s triangle, way for 0 heads, 0 ways for head, 5 ways for 2 heads, and 20 ways for heads. We want the 20. There are 2 0 = 02 ways total, so the probability is = You won t have to reduce your fractions on the tests or quizzes, but you can, if you want.