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Chapter 5: Probability: What are the Chances? Section 5.2 Probability Rules pierced pierced ears. So, (a) Each student is equally likelyto be chosen. 103 students have Considerthe exampleonSuppose page303.we choose at astudent Whenprobabilities findinginvolvingtwoevents, a two  random.Find the probabilitythatthe student the samplespacein away thatmakes probabilitycalculations easier. Two - Way Tables and Probability Define events P( pierced pierced ears) = A : is maleand (c) (b) (a) P ( B is amaleorhas pierced ears. is amalewithpierced ears. hasears. pierced B ) = 103/178. : has piercedears. - way tablecan display way

Probability Rules Probability + (b) We want to find P(male and pierced ears), that is, P(A and B). Look at the intersection of the “Male” row and “Yes” column. There are 19 males with pierced ears. So, P(A and B) = 19/178.

(c) We want to find P(male or pierced ears), that is, P(A or B). There are 90 males in the class and 103 individuals with pierced ears. However, 19 males have pierced ears – don’t count them twice! P(A or B) = (19 + 71 + 84)/178. So, P(A or B) = 174/178 The Note, the previous example illustrates the fact that can’t we use  events have no outcomes in common. the addition rule for mutually exclusive events unless the Two Venn diagram - Way Tables and Probability below below illustrates why.

Probability Rules Probability + If P A P(AP(A) +P(B) B) = U General Addition Rulefor Two Events ( A and or B B ) = from from are any two events resulting resulting events two any are process P ( some chance chance some A ) + ) or P , then ( B ) – – P P(A ∩B) ( A and B )

Probability Rules Probability + + Checkpoint  A standard deck of playing cards (with removed) consists of 52 cards in four suits – , , and . Each suit has 13 cards, with denominations , 2, 3, 4, 5, 6, 7, 8, 9, 10, , , and . The jack, queen, and king are referred to as “face cards.” Imagine that we shuffle the deck thoroughly and deal one card. Let’s define events A: getting a and B: getting a heart.

1) Make a two-way table that displays the sample space.

Face Card Non-face card Heart 3 10 Non-heart 9 30 + Checkpoint 2) Find P(A and B)

3 P(face card and heart)  52

3) Explain why P(A or B) ≠ P(A) + P(B). Then use the general addition rule to find P(A or B).

Since A and B are not disjoint, this is not equal to the sum of the probability of A and the probability of B. P(A or B) = P(A) + P(B) – P(A and B) = 0.423 overlap. That overlap.That is, no theyoutcomeshave in common. The events The The complement Because Because Venn diagrams have uses other in branches of  been been developed. mathematics, some standard vocabulary and notation have Venn DiagramsandProbability A and B A C areexclusive mutually (disjoint) becausetheydo not contains exactly the outcomes that are not in in not are that the outcomes exactlycontains A .

Probability Rules Probability + Hint: Hint: To keepthe symbolsrememberstraight, event The eventsunion of in in both events The intersection of events  Venn DiagramsandProbability Aor B. Aand A B. and B A ( and A ∪ B B ) is theset outcomes) is in of all either ( A ∩ B ) is thesetoutcomes) is of all ∪ for u nion nion and ∩ for i n tersection.

Probability Rules Probability + Recallonthe example gender andpierced ears. We cana Venn use  diagramto display the informationand determineprobabilities. Venn DiagramsandProbability Define events A : is male and and : is male B : has piercedears.

Probability Rules Probability + + Example – Venn Diagrams, Two- Way Tables and Probability

 According to the National Center for Health Statistics, in December 2008, 78% of U.S. households had a traditional landline telephone, 80% of households had cell phones, and 60% had both. Suppose we randomly selected a household in December 2008.

1) Make a two-way table that displays the sample space of this chance process.

Cell Phone No Cell Phone Total Landline 0.60 0.18 0.78 No Landline 0.20 0.02 0.22 Total 0.80 0.20 1.00 + 2) Construct a Venn Diagram to represent the outcomes of this chance process. + 3) Find the probability that the household has at least one of the two types of phones. To find the probability that the household has at least one of the two types of phones, we need to find the probability that the household has a landline, a cell phone, or both.

P(A U B) = P(A) + P(B) – P(A ∩ B) = 0.78 + 0.80 – 0.60 = 0.98

There is a 98% chance that the household has at least one of the two types of phones.

4) Find the probability that the household has a cell phone only. P(cell phone only)  P(Ac  B)  0.20 + Calculators & Probability Many probability problems involve simple computations that you can do on your calculator. It may be tempting to just write down your final answer without show the supporting work. Don’t do it!!! A “naked answer,” even if it’s correct, will usually earn you no credit on a free-response question. + Section 5.2 Probability Rules

Summary In this section, we learned that…

 A probability model describes chance behavior by listing the possible outcomes in the sample space S and giving the probability that each outcome occurs.

 An event is a subset of the possible outcomes in a chance process.

 For any event A, 0 ≤ P(A) ≤ 1

 P(S) = 1, where S = the sample space number of outcomes corresponding to event A  S P(A)  If all outcomes in are equally likely, total number of outcomes in sample space

 P(AC) = 1 – P(A), where AC is the complement of event A; that is, the

event that A does not happen.  + Section 5.2 Probability Rules

Summary In this section, we learned that…

 Events A and B are mutually exclusive (disjoint) if they have no outcomes in common. If A and B are disjoint, P(A or B) = P(A) + P(B).

 A two-way table or a Venn diagram can be used to display the sample space for a chance process.

 The intersection (A ∩ B) of events A and B consists of outcomes in both A and B.

 The union (A ∪ B) of events A and B consists of all outcomes in event A, event B, or both.

 The general addition rule can be used to find P(A or B):

P(A or B) = P(A) + P(B) – P(A and B)