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Unformatted text preview: 66 Chapter 3 Conditional Probability and Independence chosen bulb is not totally defective, the problem reduces to computing the prob
C P 3 ability that a bulb, chosen at random from a bin containing 10 partially defec
tive and 5 good bulbs, is good. I Lech (6/ When all outcomes are assumed to be equally likely, it is often easier to com N .\  pute a conditional probability by a consideration of the reduced sample space, as op
6 es posed to a direct application of (2.1). E “(AWE Examplgzg In the card game bridge the 52 cards are dealt out equally to 4 players—
cal ast, West, North, and South. If North and South have a total of 8 spades
ffo m ROSS among them, what is the probability that East has 3 of the remaining 5 spades? Solution Probably the easiest way to compute this is to work with the reduced
sample space. That is, given that North—South have a total of 8 spades among
their 26 cards, there remains a total of 26 cards, exactly 5 of them being spades,
to be distributed among the East—West hands. As each distribution is equally
likely, it follows that the conditional probability that East will have exactly 3
spades among his or her 13 cards is (ii) Exampl @ The organization for which Ms. Jones works is running a dinner for
tho * ‘ mployees having at least one son. If Jones is known to have two children,
what is the conditional probability that they are both boys, given that she
is invited to the dinner? Assume that the sample space S is given by
S = {(b, b), (b,g), (g, b), (g,g)} and all outcomes are equally likely [(b,g)
means, for instance, that the older child is a boy and the younger child is a girl]. z .339 Solution The knowledge that Jones has been invited to the dinner is equiva
lent to knowing that she has at least one son. Hence, letting B denote the event
that both children are boys and A the event that at least one of them is a boy,
we have that the desired probability P(BA) is given by PBA P(BA) = 1:04))
__mn_=zi=1
p({(b,b),(b,g),(g,b)}) % 3 Many readers incorrectly reason that the conditional probability of two
boys given at least one is %, as opposed to the correct 3%, since they reason that
the Jones child not attending the dinner is equally likely to be a boy or a girl.
Their mistake, however, is in assuming that these two possibilities are equally
likely. For, initially, there were 4 equally likely outcomes. Now the informa
tion that at least one child is a boy is equivalent to knowing that the outcome
is not (g, g). Hence we are left with the 3 equally likely outcomes (b, b), (b, g), (3,1
1y tc Byr In words,
to the pro
that F occ
the intern Example 2
ist
be 2
her <
in cl Solu
ever
prol: Example 2
balls
ball :
balls Solu'
ond l
7 ren 1y %, Of C( A gel ability of tl
as the mull )mputing the prob
10 partially defec
I ften easier to com
ample space, as op ually to 4 players—
: a total of 8 spades
:maining 5 spades?_ .‘k with the reduced
of 8 spades among
them being spades,
tribution is equally
will have exactly 3 inning a dinner for
) have two children,
)ys, given that she
ace S is given by
ually likely [(b, g)
nger child is a girl]. 1e dinner is equiva
B denote the event
1c of them is a boy, =1
3 probability of two
ce they reason that
) be a boy or a girl.
ibilities are equally
Now the informa
g that the outcome :omes (b, b), (b, g), Section 3.2 Conditional Probabilities 67 (g, b) thus showing that the Jones child not attending the dinner is twice as like
ly to be a girl as to be a boy. I By multiplying both sides of Equation (2.1) by P(F), we obtain
P(EF) = P(F)P(EF) (2.2) In words, Equation (2.2) states that the probability that both E and F occur is equal
to the probability that F occurs multiplied by the conditional probability of E given
that F occurred. Equation (2.2) is often quite useful in computing the probability of
the intersection of events. Example 2e. Celine is undecided as to whether to take a French course or a chem
istry course. She estimates that her probability of receiving an A grade would
be 5 in a French course, and g in a chemistry course. If Celine decides to base
her decision on the ﬂip of a fair coin, what is the probability that she gets an A
in chemistry? Solution If we let C be the event that Celine takes chemistry and A denote the
event that she receives an A in whatever course she takes, then the desired
probability is P(CA). This is calculated by using Equation (2.2) as follows: P(CA) = P(C)P(AC)
= (%)(%) =l  Examplg/f) Suppose that an urn contains 8 red balls and 4 white balls. We draw 2
ba rom the urn without replacement. If we assume that at each draw each ball in the urn is equally likely to be chosen, what is the probability that both
balls drawn are red? Solution Let R1 and R2 denote, respectively, the events that the first and sec
ond balls drawn are red. Now, given that the first ball selected is red, there are
7 remaining red balls and 4 white balls, and so P(Rlel) = 17—1. As P(R.) is clear
ly 1—82, the desired probability is P(RIRZ) = P(R1)P(R2IR1)
= (901) = —
Of course, this probability could also have been computed by
2
P(RIRZ) = — I
12
2
A generalization of Equation (2.2), which provides an expression for the prob ability of the intersection of an arbitrary number of events, is sometimes referred to
as the multiplication rule. 70 Chapter 3 Conditional Probability and Independence much weight as the event on which it is conditioned has of occurring. This is an ex tremely useful formula because its use often enables us to determine the probability of an event by first “conditioning” upon whether or not some second event has oc curred. That is, there are many instances where it is difficult to compute the proba bility of an event directly, but it is straightforward to compute it once we know whether or not some second event has occurred. We illustrate this with some examples.
/. Examplé‘g’an 1). An insurance company believes that people can be divided into
tw ses: those who are accident prone and those who are not. Their statis
tics show that an accidentprone person will have an accident at some time with
in a fixed 1year period with probability .4, whereas this probability decreases
to .2 for a nonaccidentprone person. If we assume that 30 percent of the pop
ulation is accident prone, what is the probability that a new policyholder will
have an accident within a year of purchasing a policy? Solution We shall obtain the desired probability by first conditioning upon
whether or not the policyholder is accident prone. Let A1 denote the event
that the policyholder will have an accident within a year of purchase; and let A
denote the event that the policyholder is accident prone. Hence the desired
probability, P(A1), is given by P(A1) = P(A1A)P(A) + P(A1A‘)P(AC)
= (.4)(.3) + (.2)(.7) = .26 I
Example 3a (Part 2). Suppose that a new policyholder has an accident within a year
of purchasing a policy. What is the probability that he or she is accident prone?
Solution The desired probability is P(AA,), which is given by P(AA1) = 135—3?
_ P(A)P(A1A)
' P(A1)
_ (.3)(.4) = i
_ .26 13 ' Examplé? Consider the following game played with an ordinary deck of 52 play ing rds. The cards are shufﬂed and then turned over one at a time. At any
time the player can guess that the next card to be turned over will be the ace of
spades; if it is, then the player wins. In addition, the player is said to win if the
ace of spades has not yet appeared when only one card remains and no guess
has yet been made. What is a good strategy; what is a bad strategy? Solution Every stategy has probability 1/52 of winning! To show this, we will
use induction to prove the stronger result that for an n card deck, one of whose
cards is the ace of spades, the probability of winning is 1 /n, no matter what strat ng den
spat
the
spa:
cart
prol
lattt
an r hyp
carc Thu Example
kn0'
the .
stud
m is
abili
SW6}
Solu SW6]
ansv ing. This is an ex
ne the probability
ond event has oc
>mpute the proba
we know whether
are examples. :an be divided into
:not. Their statis
at some time with
bability decreases
ercent of the pop policyholder will onditioning upon
denote the event
urchase; and let A
Ience the desired dent within a year
s accident prone? 1by y deck of 52 play
it a time. At any
will be the ace of
said to win if the
tins and no guess
ategy? show this, we will
:ck, one of whose
natter what strat .‘J. = Section 3.3 Bayes’ Formula 71 egy is employed. Since this is clearly true for n = 1, assume it to be true for an
'i n  1 card deck, and now consider an n card deck. Fix any strategy, and let p
denote the probability that this strategy guesses that the first card is the ace of
spades. Given that it does, then the player’s probability of winning is 1/n. On
the other hand, if the strategy does not guess that the first card is the ace of
spades, then the probability that the player wins is the probability that the first
. card is not the ace of spades, namely (n  1)/n, multiplied by the conditional
' probability of winning given that the first card is not the ace of spades. But this
latter conditional probability is equal to the probability of winning when using
an n — 1 card deck containing a single ace of spades; it is thus, by the induction
hypothesis, 1/ (n  1). Hence, given that the strategy does not guess the first
card, the probability of winning is n—1 1 _
n n1 l
n Thus, letting G be the event that the first card is guessed, we obtain that
P{win} = P{wian}P(G) + P{wian‘}(1 — P(G)) 1 1
;P+;(1—P) 1
— I
n Example 3c. In answering a question on a multiplechoice test, a student either knows the answer or guesses. Let p be the probability that the student knows the answer and 1 — p the probability that the student guesses. Assume that a
student who guesses at the answer will be correct with probability 1/m, where
m is the number of multiplechoice alternatives. What is the conditional prob
ability that a student knew the answer to a question, given that he or she an
swered it correctly? Solution Let C and K denote, respectively, the events that the student an
swers the question correctly and the event that he or she actually knows the
answer. Now P(KC) P(KC) = P(C) ; _ P(CIK)P(K)
;  ‘ P(CK)P(K) + P(CIK‘)P(K‘)
= P p +(1/m)(1  p) L
1+ (m—1)p 72 Chapter 3 Conditional Probability and Independence For example, if m = 5, p = %, then the probability that a student knew the an ' Example
swer to a question he or she correctly answered is g. I I‘m
rec«
Exampl. A laboratory blood test is 95 percent effective in detecting a certain diti
disease when it is, in fact, present. However, the test also yields a “false posi onlj
tive” result for 1 percent of the healthy persons tested. (That is, if a healthy tCSt
person is tested, then, with probability .01, the test result will imply he or she has alm
the disease.) If .5 percent of the population actually has the disease, what is the set
probability a person has the disease given that the test result is positive? 15,:
n t
Solution Let D be the event that the tested person has the disease and E it d.
the event that the test result is positive. The desired probability P(DE) is the
obtained by doe
P D E P(DE) dlrail
= o
. ____P<E£>_P<2>___ 31?!
P(EID)P(D) + P(EID‘)P(DC) thaI
_ (.95)(.005) dise
‘ (.95)(.005) + (.01)(.995) P”
95
— ﬂ ~ .323 Thus only 32 percent of those persons whose test results are positive actually
have the disease. As many students are often surprised at this result (as they ex
pected this figure to be much higher, since the blood test seems to be a good
one), it is probably worthwhile to present a second argument that, although less
rigorous than the preceding one, is probably more revealing. We now do so.
Since .5 percent of the population actually has the disease, it follows that,
on the average, 1 person out of every 200 tested will have it. The test will cor— rectly confirm that this person has the disease with probability .95. Thus, on N“ the average out of every 200 persons tested, the test will correctly confirm that _ tron .95 persons have the disease. On the other hand, however, out of the (on the cau: average) 199 healthy people, the test will incorrectly state that (199)(.01) of he ‘ these people have the disease. Hence, for every .95 diseased person that the test “0“ correctly states is ill, there are (on the average) (199) (.01) healthy persons that met
the test incorrectly states are ill. Hence the proportion of time that the test re sult is correct when it states that a person is ill is Example is 61 __25_ = E ~ new .95 + (199)(.01) 294 ~ 323 ' (5“, of tl Equation (3.1) is also useful when one has to reassess one’s personal probabilities ' sus; in the light of additional information. For instance, consider the following examples chai 5
atch provided that be the only match,
5), all others in the
IVC a match. Now,
her member of the a .a ity that all the oth tyAJ innocent) none of the others
re, 9999
) 1
.9 + 1"" U 'bitrary exconvict
was a nonconvict, ) was 0 = 10, then 3.4 Section 3.4 Independent Events 83 If the district attorney initially felt that the criminal was equally likely to be any
of the members of the town (c = 1) then a = 10‘6 and
1
P G M = — z .
(  ) 10.9 00917
Thus, the probability ranges from approximately nine percent when the district
attorney’s initial assumption is that all the members of the population have the
same chance of being the perpetrator, to approximately 91 percent when she as
sumes that each exconvict is 100 times more likely to be the criminal than is a
specified townsperson who is not an exconvict. I INDEPENDENT EVENTS The previous examples of this chapter show that P( E  F), the conditional probabil
ity of E given F, is not generally equal to P(E), the unconditional probability of E.
In other words, knowing that F has occurred generally changes the chances of E’s
occurrence. In the special cases where P(E  F) does in fact equal P(E), we say that
E is independent of F. That is, E is independent of F if knowledge that F has oc
curred does not change the probability that E occurs. Since P(ElF) = P(EF)/P(F), we see that E is independent ofFif P(EF) = P(E)P(F) (4.1) As Equation (4.1) is symmetric in E and F, it shows that whenever E is independent
of F, F is also independent of E. We thus have the following definition. . TWO events E and F :are'said to be independent if Equation (4.1) holds 1
"  " _'IVvo events E and F that are not independent are said to be dependent. ' Exam  Q A card is selected at random from an ordinary deck of 52 playing cards.
If ' he event that the selected card is an ace and F is the event that it is a spade, then E and F are independent. This follows because P(EF) = g,
whereas P(E) = ﬁand P(F) = 3. Example 4b. Two coins are flipped, and all 4 outcomes are assumed to be equal
1y likely. If E is the event that the first coin lands heads and F the event
that the second lands tails, then E and F are independent, since
P(EF) = P({(H,T)}) = g; whereas P(E) = P({(H,H), (H,T)}) = gand
P(F) = P({(H,T),(T,T)}) = i. I 84 Chapter 3 Conditional Probability and Independence Examp, Suppose that we toss 2 fair dice. Let E1 denote the event that the lil l or. cquival of  dice is 6 and F denote the event that the first die equals 4. Then P(E1F) = P( { (4, 2)}) = a
whereas and the re 5 1 P(E1)P(F) = (eXa) = % Th
Hence El and F are not independent. Intuitively, the reason for this is clear he . chm vcdub
cause if we are interested in the possibility of throwing a 6 (with 2 dice) we slllll bgupp
be quite happy if the ﬁrst die lands 4 (or any Of the numbers 1, 2, 3, 4, 5), for the lllcn Accee
we shall still have a possibility of getting a total of 6. On the other hand, if m We, the E. first die landed 6, we would be unhappy because we would no longer have
chance of getting a total of 6. In other words, our chance of getting a total of . . . Exam le 4
depends on the outcome of the first die; hence E1 and F cannot be independcnl p dice
Now, suppose that we let E2 be the event that the sum of the dice equal even
7. Is E2 independent of F? The answer is yes, since penc
P(E2F) =P({(4’3)}) =3i6 depe
whereas
1 1
P(E2)P(F) = (s)(a) = (a) ‘ d " lg;
ll c on <
We leave it for the reader to present the intuitive argument why the CVL _ p
that the sum of the dice equals seven is independent of the outcome on 1 assuming]
first die. following
Example 4d. If we let E denote the event that the next president is a Republican all . it" Deﬁl
F the event that there will be a major earthquake within the next year, th
most people would probably be willing to assume that E and F are indepen _ _The
ent. However, there would probably be some controversy over whether it ' '
reasonable to assume that E is independent of G, where G is the event Hi
there will be a recession within two years after the election.
We now show that if E is independent of F, then E is also independent of F ‘.
lt sh
: enlofany Proof: Assume that E and F are independent. Since E = EF U EF‘, and i I
and EF‘ are obviously mutually exclusive, we have that P(E) = P(EF) + P(EFC)
= P(E)P(F) + P(EF‘) and drug 2 does
[in the same with
r only those pairs
:nce will go up by P1 is equal to the
probability P will
i = M, N = 2M, incorrect decision
I ve want to deter
ty. We can attack
fthe set in such a
Then the original
andomly selected
.then we have es
1 if it is zero. then Section 3.4 Independent Events 95 Figure 3.3 The final example of this section illustrates this technique. Examp 4k. e complete graph having n vertices is defined to be a set of n points n 2 pair of vertices. The complete graph having 3 vertices is shown in Figure 3.3.
Suppose now that each edge in a complete graph on n vertices is to be colored
either red or blue. For a fixed integer k, a question of interest is whether there (called vertices) in the plane and the < > lines (called edges) connecting each k
is a way of coloring the edges so that no set of k vertices has all of its ( 2) con necting edges the same color. It can be shown, by a probabilistic argument,
that if n is not too large, then the answer is yes. The argument runs as follows. Suppose that each edge is, independently,
equally likely to be colored either red or blue. That is, each edge is red with
n probability %. Number the k i = 1, ..., (n) as follows: E, = {all of the connecting edges of the ith set
of k vertices are the same color} sets of k vertices and define the events E,, k
Now, since each of the < 2) connecting edges of a set of k vertices is equally likely to be either red or blue, it follows that the probability that they are all the
same color is P(E{) = 2(%)"‘*‘”’2
Therefore, since P<U E,> S 2 P(E,) (Boole’s inequality) we obtain that P<U E,), the probability that there is a set of k vertices all of whose connecting edges are similarly colored, satisfies P<L7J E1) 5 <:>(%)k(k~1)32—1 96 Chapter 3 Conditional Probability and Independence Hence if Proof:
n _ _ side inequal
(1)6)“ ”/2 I < 1 plies that P
or, equivalently, if
(2) < 2"("“)/2‘l Part (c) follt
then the probability that at least one of the (2) sets of k vertices has all of its _ P(
connecting edges the same color is less than 1. Therefore, under the preceding
condition on n and k, it follows that there is a positive probability that no set of
k vertices has all of its connecting edges the same color. But this implies that
there is at least one way of coloring the edges for which no set of k vertices has
all of its connecting edges the same color. I
REMARKS. (a) It should be noted that whereas the preceding argument estab
lished a condition on n and k that guarantees the existence of a coloring scheme sat g ,
isfying the desired property, it gives no information about how to obtain such a scheme. L‘
(Although one possibility would simply be to choose the colors at random, check to
see if the resulting coloring satisfies the property, and repeat this until it does.)
_ (b) The method of introducing probability to a problem whose statement is where the m
purely deterministic has been called the probabilistic method]f Other examples of
this method are given in Theoretical Exercise 22 and Example 2q of Chapter 7. If we c'
may be regai 35 H  IF) IS A PROBABILITY SlthnS previ Conditional probabilities satisfy all of the properties of ordinary probabilities. This
is proved by Proposition 5.1, which shows that P(E F) satisfies the three axioms of
a probability. or, equivaler Also, if wt
Q(E1E2)/Q( Since *See N. Alon, J. Spencer, and P. Erdos, The Probabilistic Method (New York: John Wiley & Son
Inc., 1992). 98 Chapter 3 Conditional Probability and Independence we see that Equation (5.1) is equivalent to n su
P(EllF) = P(EIIEZF)P(E2IF) + P(EIIEEF)P(E§F) m
' 1 tt‘
Exampl, Consider Example 3a, which is concerned with an insurance compa ' e 1
my . believes that people can be divided into two distinct classes—those who '
are accident prone and those who are not. During any given year an accident Clea
prone person will have an accident with probability .4, whereas the correspon first
ding figure for a nonaccidentprone person is .2. What is the conditional ing1
probability that a new policyholder will have an accident in his or her second all c
year of policy ownership, given that the policyholder has had an accident in the ed c
first year? '
Solution If we let A be the event that the policyholder is accident prone and _ 
we let Ai,i = 1, 2, be the event that he or she has had an accident in the ith __ ' AS 1
year, then the desired probability P(AzlAl) may be obtained by conditioning .. P (f
on whether or not the policyholder is accident prone, as follows:
P(AZIAI) = P(AzlAA1)P(AlA1) + P(AzlACA1)P(A€lA1)
Now, I we 1
P A A P A A P A
MIA”: (1)= (1 ) <>
PM» PM» No.
 However, P(A) is assumed to equal 13—0, and it was shown in Example 3a that 3 _. P(L
P(Al) = .26. Hence leas
_ (.4)<.3) _ 6 P‘e‘
P(A'A‘) ' .26 _ 13 '
and thus Thu
7
P(ACIA1)= 1  P(AIA1)= _
13
Solr
Since P(A2AA1) = .4 and P(AZIA‘AI) = .2, we see that
P(AZIAI) = (4% + (2% z .29 u
and The next example deals with a problem in the theory of runs. Example 5b. Independent trials, each resulting in a success with probability p or a
failure with probability q = 1  p, are performed. We are interested in com _
puting the probability that a run of n consecutive successes occurs before a run _ and
of m consecutive failures. ' Solution Let E be the event that a run of n consecutive successes occurs be
fore a run of m consecutive failures. To obtain P(E), we start by conditioning
on the outcome of the first trial. That is, letting H denote the event that the
first trial results in a success, we obtain P(E) = pP(EH) + qP(EIH‘) (52) 102 Chapter 3 Conditional Probability and Independence The notion of conditional independence can easily be extended to more than
two events and this extension is left as an exercise. The reader should note that the concept of conditional independence was im
plicitly employed in Example 5a, where it was implicitly assumed that the events that
a policyholder had an accident in his or her ith year, i = 1, 2, . .. , were conditionally
independent given whether or not the person was accident prone. [This was used to
evaluate P(AZIAA1) and P(A2A“A,) as, respectively, .4 and .2.] The following ex
ample, sometimes referred to as Laplace’s rule of succession, further illustrates the
concept of conditional independence Exampé 5d. l Laplace’s mle of succession. There are k + 1 coins in a box. The ith co 111, when ﬂipped, turn up heads with probability i/k, i = 0, 1,, k. A
coin is randomly selected from the box and is then repeatedly ﬂipped. If the first
n ﬂips all result in heads, what is the conditional probability that the (n + 1)st
ﬂip will do likewise? Solution Let C, denote the event that the ith coin is initially selected, 1' = 0,
1,... , k; let F,l denote the event that the first n ﬂips all result in heads; and let
H be the event that the (n + 1)st ﬂip is a head. The desired probability,
P(H  F"), is now obtained as follows: P(HIFn) = $P(HFnci)P(CiFn) Now, given that the ith coin is selected, it is reasonable to assume that the
outcomes will be conditionally independent, with each one resulting in a head
with probability i/ k. Hence P(HIFnci) = P(HlCz) = f
Also, P(CiFn) P(Fn) .P(FnC.)P(C.)
2P<F.1€;)P<C,) = (i/k)"[1/(k + 1)]
2(i/k)"[1/<k + 1)] P(Cian) = Hence Bul SUMMARY For even
noted by The iden is known
A 11 It can be
P(l shows th:
comes its
evidence Let
ple space is known
then Bayi
potheses ided to more than pendence was im
hat the events that
vere conditionally [This was used to
The following ex
her illustrates the sin a box. The ith
i= 0,1,...,k. A
:"lipped. If the first
that the (n + 1)st ly selected,i = O,
t in heads; and let
sired probability, 0 assume that the
esulting in a head Summary 103 But if k is large, we can use the integral approximations 1k in+1 1n+1 1
ki=0(k) ~‘/0x (ix—n+2, i*<i>"~/‘m= 1
kj=0 k 0 ”+1 n+1
n+2 so, for k large, P(Han) z SUMMARY For events E and F, the conditional probability of E given that F has occurred is de
noted by P(E F) and is defined by P(EF)
P(F) P(EF) = The identity
P(E1E2"'En) = P(E1)P(E2E1)"’P(EnlEi"'En—l) is known as the multiplication rule of probability.
A useful identity is that P(E) = P(EF)P(F) + P(EF‘)P(F‘) It can be used to compute P(E) by “conditioning” on whether F occurs.
P(H )/P(H‘) is called the odds ratio of the event H. The identity P(HIE) _ P(H) P(EIH)
P(H‘E)  P(H‘)P(EH‘)
shows that when new evidence E is obtained, the value of the odds ratio of H be
comes its old value multiplied by the ratio of the conditional probability of the new
evidence when H is true to its conditional probability when H is not true.
Let F,, i = 1, . . . , n, be mutually exclusive events whose union is the entire sam
ple space. The identity
P(E  Fj)P(F,) H g P(EIE)P(E~) P(EIE) = is known as Bayes’ formula. If the events E, i = 1, , n, are competing hypotheses,
then Bayes’ formula shows how to compute the conditional probabilities of these hy
potheses when additional evidence E becomes available. ...
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