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5.2 Probability Rules Objectives SWBAT: DESCRIBE a probability model for a chance process. USE basic probability rules, including the complement rule and.

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Presentation on theme: "5.2 Probability Rules Objectives SWBAT: DESCRIBE a probability model for a chance process. USE basic probability rules, including the complement rule and."— Presentation transcript:

1 5.2 Probability Rules Objectives SWBAT: DESCRIBE a probability model for a chance process. USE basic probability rules, including the complement rule and the addition rule for mutually exclusive events. USE a two-way table or Venn diagram to MODEL a chance process and CALCULATE probabilities involving two events. USE the general addition rule to CALCULATE probabilities.

2 What is a sample space? (not to be confused with blank space…if it was a blank space you’d write your name) What is a probability model? The sample space S of a chance process is the set of all possible outcomes. A probability model is a description of some chance process that consists of two parts: a sample space S and a probability for each outcome. The sample space S of a chance process is the set of all possible outcomes. A probability model is a description of some chance process that consists of two parts: a sample space S and a probability for each outcome. Let’s say we toss a coin one time. There are only two possible outcomes: heads and tails. We write the sample space using set notation as S={H,T}. The probability for each of these outcomes is 0.50.

3 What is an event? An event is any collection of outcomes from some chance process. That is, an event is a subset of the sample space. Events are usually designated by capital letters, like A, B, C, and so on. If A is any event, we write its probability as P(A). Let’s say we roll a pair of dice. Here is our sample space: Suppose we define event A as “sum is 5.” There are 4 outcomes that result in a sum of 5. Since each outcome has probability 1/36, P(A) = 4/36. Suppose event B is defined as “sum is not 5.” What is P(B)? P(B) = 1 – 4/36 = 32/36

4 Imagine flipping a fair coin three times. Describe the probability model for this chance process and use it to find the probability of getting at least 1 head in three flips. S = {HHH, HHT, HTH, HTT, TTT, TTH, THT, THH} Each of these 8 outcomes will be equally likely and have a probability of.125. P(at least 1 head) = 7/8 =.875.

5 Summarize the five basic probability rules. What does it mean if two events are mutually exclusive? The probability of any event is a number between 0 and 1. All possible outcomes together must have probabilities whose sum is exactly 1. If all outcomes in the sample space are equally likely, the probability that event A occurs can be found using the formula The probability that an event does not occur is 1 minus the probability that the event does occur. If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities. Two events A and B are mutually exclusive (disjoint) if they have no outcomes in common and so can never occur together—that is, if P(A and B ) = 0. Example: If we randomly select a student from LHS, what is the probability that the student is both a sophomore and a senior? 0!

6 We can summarize the basic probability rules more concisely in symbolic form. For any event A, 0 ≤ P(A) ≤ 1. If S is the sample space in a probability model, P(S) = 1. In the case of equally likely outcomes, Complement rule: P(A C ) = 1 – P(A) Addition rule for mutually exclusive events: If A and B are mutually exclusive, P(A or B) = P(A) + P(B). Basic Probability Rules

7 Randomly select a student who took the 2010 AP Statistics exam and record the student’s score. Here is the probability model: a)Show this that is a legitimate probability model. All the probabilities are between 0 and 1 and the sum of the probabilities is 1, so this is a legitimate probability model. b) Find the probability that the chosen student scored 3 or better. P(3 or better)=.235+.224+.125=.584 or P(3 or better)=1-P(2 or worse)=1-(.233+.183)=1-.416=.584 c) Find the probability that the chosen student didn’t get a 1. P(not 1)= 1-P(1)=1-.233=.767 or P(not 1)=.183+.235+.224+.125=.767

8 What is the general addition rule? Is it on the formula sheet? What if the events are mutually exclusive? If A and B are any two events resulting from some chance process, then P(A or B) = P(A) + P(B) – P(A and B) General Addition Rule for Two Events When finding probabilities involving two events, a two-way table can display the sample space in a way that makes probability calculations easier. Suppose we choose a student at random. Find the probability that the student a)has pierced ears P(pierced ears) = 103/178 b) is a male with pierced ears P(male and pierced ears) = 19/178 c) is a male or has pierced ears P(male or pierced ears) = 90/178 + 103/178 – 19/178 = 174/178 We need to subtract the 19 so that we do not double count it. Alternative method: P(male or pierced ears) = 71/178 + 19/178 + 84/178 = 174/178

9 On the formula sheet:

10 Who owns a home? What is the relationship between educational achievement and home ownership? A random sample of 500 U.S. adults was selected. Each member of the sample was identified as a high school graduate (or not) and as a home owner (or not). Overall, 340 were homeowners, 310 were high school graduates, and 221 were both homeowners and high school graduates. a) Create a two-way table for the data. Suppose we chose a member of the sample at random. Find the probability that the member b) Is a high school graduate P(graduate) = 310/500 c) Is a high school graduate and owns a home P(graduate and homeowner) = 221/500 d) Is a high school graduate or owns a home P(graduate or homeowner) = 119/500 + 221/500 + 89/500 = 429/500 or P(graduate or homework) = 310/500 + 340/500 – 221/500 = 429/500 Note: Show work. You must AT LEAST show the fraction.

11 According to the National Center for Health Statistics, in December 2012, 60% of US households had a traditional landline telephone, 89% of households had cell phones, and 51% had both. Suppose we randomly selected a household in December 2012. a)Make a two-way table to displays the sample space of this chance process. b)Construct a Venn diagram to represent the outcomes of this chance process. c)Find the probability that the household has at least one of the two types of phones P(cell or land) =.09+.51+.38 =.98 d) Find the probability that the household has neither type of phone P(no land and no cell) =.02 e) Find the probability the household has a cell phone only. P(only cell) =.38


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