288 Section 13.3 Probability Probability is a measure of how likely an event will occur. When the weather forecaster says that there will be a 50% chance of rain this afternoon, the probability that it will rain is ½. Although we cannot say with certainty that it will rain this afternoon, we can say that if the same weather patterns were repeated over and over again, we would see that it would rain about half of the time. Objective #1: Constructing a Probability Model. If we roll a fair six-sided die, we cannot be certain which number will be on top, but if we repeatedly rolled the die and recorded the number, we would see that each number would be on top about once every six times. The set of all possible results of an experiment is called the sample space. In our die example, our experiment would be the roll of one die, so the sample space would be the set of numbers that could be on top which are {1, 2, 3, 4, 5, 6}. An outcome is the result of performing one experiment (i.e., one die roll) so each element of the sample space is a possible outcome. For each outcome, we can assign the probability that the outcome would occur. Since we are rolling a fair die, the probability of each outcome is 1/6. Thus, the probability of rolling a five is 1/6. We can denote this as P(5) = 1/6. For all of the possible outcomes, we can construct a probability model: P(1) = 1/6 P(2) = 1/6 P(3) = 1/6 P(4) = 1/6 P(5) = 1/6 P(6) = 1/6 Notice that the probability each outcome is greater than or equal 0 and the sum of the probabilities of all the outcomes is 1. These are two important conditions of a probability model. Definition: Probability Model Let S = {e 1, e 2,, e n } be the sample space of a probability model where e 1, e 2,, e n are the possible outcomes of an experiment and let P(e 1 ), P(e 2 ),, P(e n ) be the corresponding probabilities of the outcomes. Then, the probabilities of the outcomes must satisfy both of the following conditions: 1) P(e i ) 0 for all natural numbers i where 1 i n. 2) P(e 1 ) + P(e 2 ) + + P(e n ) = 1 Note that the two conditions together also imply that the probability of each outcome has to be between 0 and 1 inclusively (i.e., 0 P(e i ) 1)
Determine whether the following are probability models: Ex. 1a Ex. 1b Ex. 1c Outcome Probability Outcome Probability Outcome Probability A 0.4 1 0.25 Red 0.5 B 0.3 2 0.50 Blue 0.3 C 0.1 3 0.10 Green 0.1 D 0.4 4 0.15 Black 0.3 289 a) Since the sum of the probabilities is equal to 1.2 and not 1, then this is not a probability model. b) Since the probability of each outcome is greater than zero and the sum of the probabilities is equal to 1, then this is a probability model. c) Since P(Green) < 0, then this is not a probability model. Construct a probability model for the following: Ex. 2 A fair coin is tossed twice. Construct a probability model. First, let's construct the sample space by making a tree diagram. H will denote heads and T will denote tails: H HH Thus, S = {HH, HT, TH, TT} and H since each outcome is equally likely, T HT then the probability of each outcome is ¼. This means that P(HH) = ¼, H TH P(HT) = ¼, P(TH) = ¼, and P(TT) = ¼. T T TT Ex. 3 Suppose now the coin from the last example is not fair in that heads is four times more likely to occur than tails. If this coin is tossed twice, construct a probability model. Even though the sample space is the same, the probabilities are not. Let x = the probability that tails occurs on a single toss of the coin. The probability that heads occurs will be 4x. So, P(T) + P(H) = x + 4x = 1 5x = 1 x = 1 5
290 This means that P(T) = 1 5 and P(H) = 4 5 To construct the probability model of tossing the coin twice, we will need to multiply the probability of the first outcome times the probability of the second outcome: P(HH) = P(H) P(H) = 4 5 4 5 = 16 25 P(HT) = P(H) P(T) = 4 5 1 5 = 4 25 P(TH) = P(T) P(H) = 1 5 4 5 = 4 25 P(TT) = P(T) P(T) = 1 5 1 5 = 1 25 Definition: Event An event E is any subset of the sample space of an experiment. The probability of an event E, E, is equal to the sum probabilities of the outcomes in E. If we are tossing a fair coin twice, the event of tossing one head and one tail would be the sum of P(HT) + P(TH) = ¼ + ¼ = ½. When we discuss the probability of event E, we are asking what is the likelihood that event E will happen. If E =, then P(E) = 0 and if E = S, then P(E) = 1. Objective #2: Calculating the Probability of Equally Likely Outcomes. When each outcome in a sample space has the same probability of occurring, the experiment is said to have equally likely outcomes. Theorem: Probability for Equally Likely Outcomes Let S be the sample space of an experiment that has r equally likely outcomes. If the number of ways event E can occur is m, then the probability of E is: P(E) = Number of ways that E can occur Number of possible outcomes = n(e) n(s) = m r Find the following: Ex. 4 Calculate the probability that a family with three children will have all boys or all girls. By the counting principle, there are 2 3 = 8 possible permutations. Thus, the sample space will be: S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}
The event E of having all boys or all girls is E = {BBB, GGG}. Since the outcomes are equally likely, then P(E) = n(e) n(s) = 2 8 = 1 4. 291 Ex. 5 Suppose a fair die is rolled. Let E be the event "rolled an even number," and let F be the event "rolled a 5 or 6." a) Write the event E and F. What is n(e F)? b) Write the event E or F. What is n(e F)? c) Compute P(E) and compute P(F). d) Compute P(E F). e) Compute P(E F)? Since the sample space is {1, 2, 3, 4, 5, 6}, then n(s) = 6. The event E: "rolled an even number" is {2, 4, 6} and event F: "rolled a 5 or 6" is {5, 6}. Thus, n(e) = 3 and n(f) = 2. a) Since the word "and" means intersection, then the event E and F is E F = {2, 4, 6} {5, 6} = {6}. Hence, n(e F) = 1. b) Since the word "or" means union, then the event E and F is E F = {2, 4, 6} {5, 6} = {2, 4, 5, 6}. Hence, n(e F) = 4. c) P(E) = n(e) n(s) = 3 6 = 1 2 P(F) = n(f) n(s) = 2 6 = 1 3 d) P(E F) = n(e F) = 1 n(s) 6 e) P(E F) = n(e F) = 4 n(s) 6 = 2 3 Notice that if we add P(E) and P(F) together and then subtract P(E F), we get P(E F): P(E) + P(F) P(E F) = 1 2 + 1 3 1 6 = 3 6 + 2 6 1 6 = 4 6 = 2 = P(E F) 3 The reason this works is that when we add the probabilities of event E and F together, we have counted the outcomes that are in both events E and F twice. By subtracting P(E F) from the results, we have removed the duplication of the count. This introduces the next objective.
292 Objective #3: Finding the Probability of the Union of Two Events. Theorem: Probability of the Union of Two Events For any two events E and F of the same experiment, then P(E F) = P(E) + P(F) P(E F) Solve the following: Ex. 6 If P(E) = 0.25, P(F) = 0.45 and P(E F) = 0.1, find a) P(E F). b) The probability of E, but not F. c) The probability of F, but not E. d) The probability of neither E nor F. a) P(E F) = P(E) + P(F) P(E F) = 0.25 + 0.45 0.1 = 0.6 b) We want to calculate the probability that E occurred and not F occurred. The will be the probability of E minus the probability of E F. P(E) P(E F) = 0.25 0.1 = 0.15 c) We want to calculate the probability that F occurred and not E occurred. The will be the probability of F minus the probability of E F. P(F) P(E F) = 0.45 0.1 = 0.35 d) The probability that neither E nor F occurred is the probability of the entire sample space minus the probability of E F: P(S) P(E F) = 1 0.6 = 0.4 If the events E and F have no outcomes in common, then E F = and we say the two events are mutually exclusive. Theorem: Mutually Exclusive Events If events E and F are mutually exclusive events of the same sample space, then P(E F) = P(E) + P(F) Solve the following: Ex. 7 If P(E) = 0.35, P(F) = 0.55 and E and F are mutually exclusive, find P(E F). Since E and F are mutually exclusive, then P(E F) = P(E) + P(F) = 0.35 + 0.55 = 0.9 Objective #4: Using the Complement Rule to Find Probabilities.
293 Sometimes when answering a question, it is easier to answer the opposite question first. Consider the following question: What is the minimum number of games needed to determine the champion of a sixty-four-team tournament? Since you have one winner, you would then need to match teams up in games to determine how many games you will need for one winner. If, instead of thinking of having one winner, you thought about the opposite; how many losers do you need? The answer is 63 losers so you would need a minimum of 63 games to determine the champion. This idea is also useful in probability. It is sometimes easier to find the probability of something not occurring and subtract the answer from one than it is the find the probability of the event directly. In order to do this, we need to define the complement of an event. Definition: Complement of an Event Let E be an event of the sample space S. The complement of E, denoted E, is the set of all outcomes in the sample space S that are not outcomes in E. Since E and E are mutually exclusive, then they have the following properties: 1) E E = 2) S = E E 3) P(S) = 1 = P(E E) = P(E) + P( E) P( E) = 1 P(E) Theorem: Computing Probabilities of Complementary Events Let E be an event of the sample space S and let E be the complement of E, then P( E) = 1 P(E) Solve the following: Ex. 8 The percentage of students with student loans that defaulted on their federal student loans within the first two years of their first payment was 9.1% in 2011. If a student from this group is selected at random, what is the probability that the student did not default in his/her loans with the first two years of repayment? Since the complement of defaulting is not defaulting, then P(Not Defaulting) = 1 P(Defaulting) = 1 0.091 = 0.909. Thus, there is a 90.9% that the student selected did not default.
294 Ex. 9 What is the probability that in a class of 20 people, at least two people will have the same birthday? Assume that there are 365 days in a year. When we say at least two people will have the same birthday, we mean two or more. That can get very complicated to figure out since we would have add the probability that two people have the same birthday plus the probability that three people have the same birthday, etc all the way up to the probability that all 20 had the same birthday. But, the complement of at least two people having the same birthday is no one would have the same birthday. We can calculate that fairly easily. But, subtracting that answer from one, we will then get the answer to the question. Let E be the event that no two people have the same birthday. Then, n(e) = P(365, 20) = 365! (365 20)! = 365! 345! = 1.03669 10 51 and n(s) = 365 20 = 1.761397 10 51 Thus, P(E) = n(e) n(s) = 1.03669... 1051 1.761397... 10 = 365 364 346 51 = 0.58856 Finally, the probability that at least two people have the same birthday is: P( E) = 1 P(E) = 1 0.58856 = 0.411438 0.411