5.2 probability rules objectives swbat: describe a probability model for a chance process. use basic...

5.2 Probability RulesObjectives

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.

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.

• 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.

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

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.

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!

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(AC) = 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

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=.584orP(3 or better)=1-P(2 or worse)=1-(.233+.183)=1-.416=.584c) Find the probability that the chosen student didn’t get a 1.P(not 1)= 1-P(1)=1-.233=.767orP(not 1)= .183+.235+.224+.125=.767

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, thenP(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 studenta) has pierced earsP(pierced ears) = 103/178b) is a male with pierced earsP(male and pierced ears) = 19/178c) is a male or has pierced earsP(male or pierced ears) = 90/178 + 103/178 – 19/178 = 174/178We 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

On the formula sheet:

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 memberb) Is a high school graduateP(graduate) = 310/500c) Is a high school graduate and owns a homeP(graduate and homeowner) = 221/500d) Is a high school graduate or owns a homeP(graduate or homeowner) = 119/500 + 221/500 + 89/500 = 429/500or P(graduate or homework) = 310/500 + 340/500 – 221/500 = 429/500

Note: Show work. You must AT LEAST show the fraction.

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 = .98d) Find the probability that the household has neither type of phoneP(no land and no cell) = .02e) Find the probability the household has a cell phone only.P(only cell) = .38