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Chapter 5: Probability 95
Chapter 5 Multiple Choice Practice1
;1,,n/vt/\Directions. ldentify the chaice that best completes
the statement or answers the question. Check your answersand 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
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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 'l-T-)
-\ P i r)*. rt- 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\ 1/2LIS sua
(D) 1/3{E) 1t4
}L
l/!k
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$rWhat is the probability that both candies
are red?
ic)a.zzzz T, ?L?.,11 : -,lSSl(D) 0.4444 5 s t{ q(E) 0.8000 t
I
7. An event A will occur with probability 0.5. An event B
willoccur with probability 0.6. The probability that
,. ?:yl un.d ? witt occur is 0.1. d_*n T-^dtf*Jr^-{ ; {i./+.} .
.flJlj . ,f & * *{ 4$-- 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--*-:------. " *,-jEvent 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
ir iti lni -'
9. At Lakeville South High School,60%o of students have
high-speed internet access, 30% have a mobilecomputing 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%
' ., I '.ll,t, il
- 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
i.! . ....iri-'
I t-*-iit
i
readings if all suspects plead innocent and are telting the
truth is:
' -'St,r r-s '$f .' :"''r {T .! l^ *i
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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 ofinformation: 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.
*"
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 exclusiveevenfs
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?
TE..e o'*r *rJ-l -:;;;:
'*'* * 5: U:t.T.*t'*-t;""*I{'
f -t* L. str\ fsptfr i1ai & [}#il !a+'r ''1* " t.}
-1 i": e'-
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
t':/i';" 'r-.,1ri. r-e .-' , *rt'.r.t . ., 1
3. What is P(Ac)?
., -^ gl :?.r.:'ri6-j"h -'i ',--
nAl
, ,r.iL :t" '---\ .-l' t' '
:.: #, Q ?.-3,,'r
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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 organizeand 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.c-
T'l* t
2. Use a Venn diagram to display the sample space.
3. Find P(A v B). Show your work.
F {k,) 15\ = fiA-{ ro l!) -'::- u'/- - + \i/, -/s> o5 +
lvt,1f )'Jt-'
5* a, 3r: ??
? {fr n8}l)ir z-
t1lb
sbraqq
I
$l,.t
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Chapter 5: Probability 91
After You Read: Check for Understanding
Concept l: Conditional Probability and lndependenceA conditional
probability describes the chance that an event will occur given
that another event isalready known to have happened. To note that
we are dealing with a conditional probability, we usethe symbol I
to mean "given that." For example, suppose we draw one card from a
shuffled deck of 52playing cards. We could write "the probability
that the card is an ace given that it is a red card as P(aceI red).
Building on the concept of conditional probabilities, we say that
when knowing that one eventhas occurred has no effect on the
probability of another event occurring, the events are
independent.That is, events A and B are independent if P(A I B) =
P(A)and P(B lA) = P(B). Note that the events "getan ace" and "get a
red card" described earlier are independent - if you know a
randomly selected cardis red, the probability it is an ace is 2126.
This is equal the same as the probability a randomly selectedcard
(regardless of colo$ 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.
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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.
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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{ '}DPtsr+,T+rb sJ
(, q>)(, uq),5tb)
g,t4 S f 5
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262 Strive for a 5: Preparing for the AP@ Statistics
Examination
Chapter 5: ProbabilitytN tc
,S *MU o
A,A
U M N
M D,S
L B D
P D N T E R S E c T I o NL I M R T D
,EV E N T U P S E
X S L L 'b o PC
,hR o B A B I L I T N E
L A N T C S A N
U C A J L D
'h A E S E o T o EI N I I
,AA N D o 'l(/l
V o N T oE
,UN o N T D
't R E EL
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]
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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 /[solT5.11 (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
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