Chapter 5: Probability 95

Chapter 5 Multiple Choice Practice1

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Directions. ldentify the chaice that best completes the statement or answers the question. Check your answers

and note your pertormance when you are finished.

1. The probability that you will win a prize in a carnival game is about 1/7. During the last nine attempts, youhave failed to win. You decide to give it one last shot. Assuming the outcomes are independent from gameto game, the probability that you will win is:

(A) 1/7

{B) (1/7) - (1/7)'(C) (1t7) + (1/7)'g

(D) 1/10(E) 7/10

2. A friend has placed alarge number of plastic disks in a hat and invited you to select one at random. Heinforms you that half are red and half are blue. lf you draw a disk, record the color, replace it, and repeat100 times, which of the following is true?(A) lt is unlikely you wiil choose red more than 50 times.(B) lf you draw 10 blue disks in a row, it is more likely you will draw a red on the next try.(C) The overall proportion of red disks drawn should be close to 0.50.(D) The chance that the 100ih draw will be red depends on the results of the first 99 draws.(E) All of the above are true.

3. The two-way table below gives information on males and females at a high school and their preferredmusic format.

CD mp3 Vinyl Totals

Males 146 106 48 300Females 146 64 4A 2N

Totals 292 170 88 550

You select one student frorn ihis group at random. Which of the following statement is true about theevents "prefers vinyl" and "Male"?(A)Theeventsaremutuallyexclusiveandindependent.(B) The events are not mutually exclusive but they are independent.(C) The events are mutually exclusive, but they are noi independent.

_(Dj The events are not mutually exclusive, nor are they independent.(E) The events are independent, but we do not have enough information to determine if they are

mutually exclusive.

4. Peopie with type O-negative blood are universal donors. That is, any patient can receive a transfusion ofO-negative biood. An$ 7.2% of the American population has O-negative blood. If '10 people appear atrandom to give blood, what is the probability that at least 1 of them is a universal donor?(A) 0(B) 0.280(c) 0.526(D) 0.720(E) 1

96 Strive for a 5: Preparing for the AP@ Statistics Examination

5. A die is loaded so that the number 6 comes up three times as often as any other number. What is theprobability of rolling a 4, 5, or 6? h-( ) 2/3 Pl.l) +f

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t- Y' {- )r' 'i't r, lt- 'r'{- * Y&- +

f"t} { r/si*/d'e}-"t'"* "l \d "{ 3 X. =' *PX

.n, + l* * lF = !6 :' '

(B\ 1/2

LIS sua(D) 1/3

{E) 1t4

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6. You draw two candies at random from a bag that has 20,red, 19glepn, 15 orange, and 5 blue candies

Y*lre-Vl-f'-eelgg.e{9$rWhat is the probability that both candies are red?

ic)a.zzzz T, ?L?.,11 : -,lSSl(D) 0.4444 5 s t{ q(E) 0.8000 t I

7. An event A will occur with probability 0.5. An event B willoccur with probability 0.6. The probability that

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witt occur is 0.1. d_*n T-^dtf*Jr^-{ ; {i./+.}

. .flJlj . ,f & * *{ 4$-- Events A and B are independent. d '' r. q'\ ,', f,\ J tL+lef EventsAaldBaremutuallyexclusive. f-^***s_ -)

!! !''' L- *-'/ril 'irrs ^ qr r"-- \(Q Either A or B always occurs.-'** - !r , /4Tf;F. i

,p| Events L.und.B are.complementary. ' i L.J-*- :.s : , L4 .r .{ +. 5 : i(E) None of the above is correct. .t--*-:------. " *,-j

Event A occurs with probability 0.8. The conditional probability that event B occurs, given that A occurs, is0.5. The probability that both A and B occur is:

,e kj =- , f"" f { fi ip} = ,.5 Pi.* nA}:: I(A) 0,3(Ff o.a(c) 0.625(D) 0.8(E) 1.0

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9. At Lakeville South High School,60%o of students have high-speed internet access, 30% have a mobile

computing device, and 2Ao/o have both. The proportion of students that have neither high-speed internet

access nor a mobile computing dev_ice-is:

(A) o% r "-,----* I(BJ 10% i ,''"f ' ] ;, {. l

flOrgOX , r , 'i -.i .1 .' ,

(D) B0% i_---)_-_ -_ : ,3J(E) e0%

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- i ; br\'{

, i".r*.""4 n\, i v : * i, . ':-j

probability of(A) 0.40e(B) 0.735(c] 0.00001

(Doo.sgr(E) 0.geeee

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i

readings if all suspects plead innocent and are telting the truth is:

' -'St,r r-s '$f .' :"''r {T .! l^ *i<^,: - 'rn1 . ,. .' , d i,,,-if 1<.+;,1 f.)5=" (en f< a .-i .i v--,

10. Experience has shown that a certain lie detector willshow ?gg-sjlryg*r-qggjlE Otc_qles a lie) 10% of

Jh*lfg_yhgl€ person is telling the truth and 95% of the tifie-w-hen:a peison EliliiglSuF0asetn-m a

nenfien-#mdgffisusirdfitii i5 subTefieh to a lie detector test regarding a recent one-person crime. The

s8 strive for a 5: Preparing for the AP@ Statistics Examination

After You Read: Check for Understanding

Concept 1: Probabitity Modets and the Basic Bules of ProbabilityChance behavior can be described using a probability model. This model provides two pieces of

information: a list of possible outcomes (samplespace) and the likelihood of each outcome. Bydescribing

chance behavior with a probability model, we can find the probabitity of an event-a particular outcome

or collection of outcomes. probability models must obey some basic rules of probability:

,'''"o Far any event A, 0 < P(4 < 1 .

' r lf S is the sample space in a probability model, P(S) = 1', o ln the case of equaily likely outcomes, P(A) = {# outcomes in event A\ / (# outcomes in Q.

:' 'P(Ac)=1-P{ALj--" . lf A and B are mutually exclusive events, P(A or B) = P(A) + P(B).

After this section, you should be able to describe a probability model for chance behavior and apply

the basic probability rules to answer questions about events.

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checkfor understanding:

-t can describe a probability modelfor a chanc-e process and

-lcan use basic probability /ules s'uch as the complement rule and addition ru{e for mutually exclusive

evenfs

Consider drawing a card from a shuf{led fair deck of 52 playing cards'

1. How many possible outcomes are in the sample space for this chance process? What's the

probability for each outcome?

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f -t* L. str\ fsptfr i1ai & [}#il !a+'r ''1* " t.}

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Define the following events:A: the card drawn is an AceB; the card drawn is a heaft

2. Find P(A)and P(B).'^.' Ll

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t'.r.t . ., 1

3. What is P(Ac)?

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Chapter 5: Probability 89

Concept 2 Two-Way Tables and Venn DiagramsOften we'll need to find probabilities involving two events. ln these cases, it may be hetpfulto organize

and display the sample space using a two-way table or Venn diagram. This can be especially helpfulwhen two events are not mutually exclusive. When dealing with two events A and B, it is importantto be able to describe the union (or collection of all outcomes in A, 8, or both) and the intersec]ion(the collection ql upon the

wi\Dastc rutesLtq*"'.'" v-.

&:ffi;"tiontohelpusfindtheprobabilityoftwoeventsthatarenotmutuallyexclusive.

C. llA and B aretwo events , P(A v B) = P(A) + P(B) - P(A^ Bl.J

Check for Understanding: _l can use a Venn diagram to model a chance processinvolving two events, I can find the probability of an event using a two-way table, and

-l can use the general addition rule to calculate P(A U B)

Consider drawing a card from a shuffled fair deck of 52 playing cards.Define the following events:A: the card drawn is an AceB; the card drawn is a heart

1. Use a two way table to display the sample space.

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2. Use a Venn diagram to display the sample space.

3. Find P(A v B). Show your work.

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Chapter 5: Probability 91

After You Read: Check for Understanding

Concept l: Conditional Probability and lndependenceA conditional probability describes the chance that an event will occur given that another event isalready known to have happened. To note that we are dealing with a conditional probability, we usethe symbol I to mean "given that." For example, suppose we draw one card from a shuffled deck of 52playing cards. We could write "the probability that the card is an ace given that it is a red card as P(ace

I red). Building on the concept of conditional probabilities, we say that when knowing that one eventhas occurred has no effect on the probability of another event occurring, the events are independent.That is, events A and B are independent if P(A I B) = P(A)and P(B lA) = P(B). Note that the events "getan ace" and "get a red card" described earlier are independent - if you know a randomly selected cardis red, the probability it is an ace is 2126. This is equal the same as the probability a randomly selectedcard (regardless of colo$ is an ace. That is, 2/26 = 4/52. Likewise, if you know a randomly selected cardis an ace, the probability it is red is 1/2. This is the same as the probability a randomly selected card(regardless of value) is red. Therefore, the two events are independent.

Check for Understanding: _ I can compute conditional probabilities and _ I can determinewhether two events are independent

ls there a relationship between gender and candy preference? Suppose 200 high school studentswere asked to complete a survey about their favorite candies. The table below shows the gender ofeach student and their favorite candy.

Male Female TotalSkittles 80 60 140M&M's 40 2A 60Total 12O 80 2AO

Define A to be the event that a randomly selected student is male and B to be the event that arandomly selected student likes Skiftles. Are the events A and B independent? Justify your answer.

92 Strive for a 5: Preparing for the AP@ Statistics Examination

Concept 2; Tree Diagrams and the Multiplication RuleWhen chance behavior involves a sequence of events, we can model it using a tree diagram. A treediagram provides a branch for each outcome of an event along with the associated probabilities of thoseoutcomes. Successive branches represent particular sequences of outcomes. To find the probability ofan event, we multiply the probabilities on the branches that make up the event.This leads us to the general multiplication rule: P(A n B) = P(A) . P(B I A).

"lf A and B are independent, the probability that both events_occur.is P(A n B) = P{A) . P(B),"

Check for Understanding: _ I can use a tree diagram to describe chance behavior and_ I can use the general multiplication rule to solve probability questrbns

A study of high school juniors in three districts - Lakeville, Sheboygan, and Omaha - was conductedto determine enrollment trends in AP mathematics courses-Calculus or Statistics. 42Yo of studentsin the study came from Lakeville, ST%6 came from Sheboygan, and the rest came from Omaha. lnLakeville, 64% o'f juniors took Statistics and the rest took Calculus. 58% of juniors in Sheboygan and49o/o of juniors in Omaha took Statistics while the rest took Calculus in each district. No juniors tookboth Statistics and Calculus. Describe this situation using a tree diagram and find the probability thata randomly selected student from in the study took Statistics.

Chapter 5: Probability

Concept 3: Calculating Conditional ProbabilitiesBy rearranging the terms in the general multiplication rule, we can determine a rule for conditional

'using a two-way table, Venn diagram, or tree diagram. However, the formula can also be used if youknow the appropriate probabilities in the situation.

Check for Understanding: _l can compute conditional probabilities

Consider the situation from Concept 2.

Find P(student is from Lakeville ltook Statistics). = I Cr* heu ,_l I .. A 3+"&r{ '}D

Ptsr+,T+rb sJ

(, q>)(, uq)

,5tb)g,t4 S f 5

262 Strive for a 5: Preparing for the AP@ Statistics Examination

Chapter 5: ProbabilitytN tc

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P D N T E R S E c T I o N

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R o B A B I L I T N E

L A N T C S A N

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Across8. The collection of outcomes that occur in both of

two events. INTERSECTION]9. A collection of outcomes from a chance process.

lEVENrl11. The propofiion of times an outcome would occur

in a very long series of repetitions.IPROBABTLTTYI

12. _Theorem can be used to find probabilitiesthat require going "backward" in a tree diagram.IBAYESI

13. ln statistics, this doesn't mean "haphazard." ltmeans "by chance." [HANDOM]

15. The collection of outcomes that occur in eitherof two events. IUNION]

16. A _ diagram can help model chancebehavior that involves a sequence of outcomes.lrREEl

Down1. The law of large states that the

proportion of times an outcome occurs in manyrepetitions will approach a single value.INUMBERSI

2.The probability that one event happens givenanother event is known to have happened.lcoNDrTroNALl

3. The set of all possible outcomes for a chanceprocess (two words). [SAMPLESPACE]

4. The probability that two events both occur canbe found using the general _ rule.IMULTTPLTCATTONI

5. P(A or B) can be found using the generalrule. [ADDITION]

6. The imitation of chance behavior, based on amodel that reflects the situation. [SIMULATION]

7.The occurrence of one event has no effect onthe chance that another event will happen.ITNDEPENDENTI

9. Another term for disjoint: MutuallyIEXCLUSTVEI

10. Two events that have no outcomes in commonand can never occur together. [DISJOINT]

AP Statistics Practice Test (page 336)

T5.1 c. Probability only tells us what happens approximately in the long run, not what will happen in theshort run.

T5.2 d. You need exactly 62 of the 100 2-digit numbers to represent the event "having heard of Coca-cola.''

T5.3 c. Add the probabilities for 3, 4 and 5 cars.

T5.4 b. AllZ-digit numbers among the first 10 are between 00 and 97 except 98.

T5.5 b. 255 of the 1000 students had a GPA below 2.

T5.6 c. There are 285 students who either have a GPA below 2,have skipped many classes or both.

T5.7 e. There are 110 students who have skipped many classes. 80 of them have a GPA below 2.

T5.8 e. If A and B are independent, then we don't know whether B has occurred if A occurred. But if Aand B are mutually exclusive, then if B has occurred then we know that A couldn't have occurred.

T5.9 b. P(woman L., never married) = P(woman)+ P(never manied) - P(woman . never married).

T5.10 c. We want P(first is picture n second is picture n third is picture) = fg.]f+)[*) = 0.0,.' IszJIs r /[sol

T5.11 (a) Since each outcome is equally likely and there are 48 outcomes, each outcome has probability

a. th... are 27 waysin which the teacher wins (all boxes above and to the right of the diagonal line48

indicating ties). So the probability that the teacher wins is 4 fq We use the fact that48

r(at,e)=P(A)+r(n)-P(AnB). From part(a), r(a)= !. Th"r"are8outcomesinwhich"you

get a3" so f (e) = -.!- urd there are 5 outcomes in which "you roll a 3" and the teacher still wins so

r(anB)=+. Puttingall ofthistogethergives P(AuB)= +.+-+=+=: (c) From part\/4ge\'4g4g4g4gg"

(b). we have P(A) =#. ,tul: * ",0 P(A n B) : a. sin..

r(a ne) =*-(#)t*)= r(a)r(e), A and B are not independent.

T5.12 (a)

r30 The Practice of Statistics /or AP*,4/e

0q

J9, -0.!

99c <qe

Probability

Defective 0.06

oK 0.54

Defective 0,09

OK

Defective 0.04

oK 0.06

0.21

(b) To get the probability that apart randomly chosen from all parts produced by this machine isdefective, add the probabilities from all branches in the tree that end in a defective paft.

P(defective) = 0.06 + 0.09 + 0.04 = 0.19. (c) First compute the conditional probabilities that the part was

produced on a parlicular machine given that it is defective P(A ldefective) = +ry:0.3158.' 0.19

P(Bldefectir.)=:?::0.4737. r(cloerective)=901:0.210s. sincethelargestofthesettuee\ ' 0.19 \ ' 0.19conditional probabilities is for machine B, given that a part is defective, it is most likely to have come

from machine B.

T5. t3 (a) p(gers cancer I smoker) - P(gets :lncer n s-moker)

= qry

= 0.32. (b) The event thar anP(smoker) 0.25

individual either smokes or gets cancer is the complement of the event that the individual neither smokesnor gets cancer. So

P(smokesurgetscancer)=1-P(doesnotsmokendoesnotgetcancer)=1-0.71 =0.29.

(c) First find the probability that one person gets cancer. From part (b) we know that

P(smokes u gets cancer) =0.29. But we also know that P(smokes) = 9.25 und

P(smokes n gets cancer) : 0.08. Now, using P(s u C) = r(s) + r(C) - P(S n C) we get

0.29=0.25+P(getscancer)-0.08. Solvingfor P(getscancer)weget0.12. Finally

P(at least one gets cancer) = 1* P(neither gets cancer) = 1- 0.88'z = 0.2256.

T5.la (a) Assign the numbers 01-17 to represent cars with out-of-state plates and the remaining 2-digitnumbers between 00 and 99 to represent other cars. Staft reading 2-digit numbers from a random numbertable until you have found two numbers between 01 and 17 (repeats are allowed). Record how many 2-

digit numbers you had to read in order to get 2 numbers between 0 I and 1 7. Repeat many times for the

simulation. (b) The first sample is 41 05 09. The numbers in bold represent cars with out-of-stateplates. In this sample it took three cars to find two with out-of-state plates. The second sample is 20 3l06 44 90 50 59 59 88 43 18 80 53 11. In this sample it took 14 cars to find two with out-of-state plates.

The third sample is 58 44 69 94 86 85 79 67 05 81 18 45 14. In this sample it took 13 cars to find twowith out-of-state plates.

Chapter 5: Probability: What Are the Chances? l3l