TEST BANK FOR A Modern Introduction to Probability and Statistics

458 Full solutions from MIPS: DO NOT DISTRIBUTE
29.1 Full solutions
2.1 Using the relation P(A [ B) = P(A)+P(B)¡P(A \ B), we obtain P(A [ B) =
2=3 + 1=6 ¡ 1=9 = 13=18:
2.2 The event “at least one of E and F occurs” is the event E [ F. Using the
second DeMorgan’s law we obtain: P(Ec \ Fc) = P((E [ F)c) = 1 ¡ P(E [ F) =
1 ¡ 3=4 = 1=4.
2.3 By additivity we have P(D) = P(Cc \ D)+P(C \ D). Hence 0:4 = P(Cc \ D)+
0:2. We see that P(Cc \ D) = 0:2. (We did not need the knowledge P(C) = 0:3!)
2.4 The event “only A occurs and not B or C” is the event fA \ Bc \ Ccg. We
then have using DeMorgan’s law and additivity
P(A \ Bc \ Cc) = P(A \ (B [ C)c) = P(A [ B [ C) ¡ P(B [ C) :
The answer is yes , because of P(B [ C) = P(B) + P(C) ¡ P(B \ C)
2.5 The crux is that B ½ A implies P(A \ B) = P(B). Using additivity we obtain
P(A) = P(A \ B)+P(A \ Bc) = P(B)+P(A n B). Hence P(A n B) = P(A)¡P(B).
2.6 a Using the relation P(A [ B) = P(A) + P(B) ¡ P(A \ B), we obtain 3=4 =
1=3 + 1=2 ¡ P(A \ B), yielding P(A \ B) = 4=12 + 6=12 ¡ 9=12 = 1=12.
2.6 b Using DeMorgan’s laws we get P(Ac [ Bc) = P((A \ B)c) = 1 ¡ P(A \ B) =
11=12.
2.7 P((A [ B) \ (A \ B)c) = 0.7.
2.8 From the rule for the probability of a union we obtain P(D1 [ D2) · P(D1) +
P(D2) = 2 ¢ 10¡6. Since D1 \ D2 is contained in both D1 and D2, we obtain
P(D1 \ D2) · minfP(D1) ; P(D2)g = 10¡6. Equality may hold in both cases: for
the union, take D1 and D2 disjoint, for the intersection, take D1 and D2 equal to
each other.
2.9 a Simply by inspection we find that
A = fTTH; THT;HTTg;B = fTTH; THT;HTT; TTTg;
C = fHHH;HHT;HTH;HTTg;D = fTTT; TTH; THT; THHg.
2.9 b Here we find that Ac = fTTT; THH;HTH;HHT;HHHg;
A [ (C \ D) = A [ ; = A;A \ Dc = fHTTg:
2.10 Cf. Exercise 2.7: the event “A or B occurs, but not both” equals C = (A[B)\
(A \ B)c Rewriting this using DeMorgan’s laws (or paraphrasing “A or B occurs,
but not both” as “A occurs but not B or B occurs but not A”), we can also write
C = (A \ Bc) [ (B \ Ac).
2.11 Let the two outcomes be called 1 and 2. Then ­ = f1; 2g, and P(1) = p; P(2) =
p2. We must have P(1) + P(2) = P(­) = 1, so p + p2 = 1. This has two solutions:
p = (¡1 +
p
5 )=2 and p = (¡1 ¡
p
5 )=2. Since we must have 0 · p · 1 only one is
allowed: p = (¡1 +
p
5 )=2.
2.12 a This is the same situation as with the three envelopes on the doormat, but
now with ten possibilities. Hence an outcome has probability 1=10! to occur.
2.12 b For the five envelopes labeled 1; 2; 3; 4; 5 there are 5! possible orders, and
for each of these there are 5! possible orders for the envelopes labeled 6; 7; 8; 9; 10.
Hence in total there are 5! ¢ 5! outcomes.
29.1 Full solutions 459
2.12 c There are 32¢5!¢5! outcomes in the event “dream draw.” Hence the probability
is 32 ¢ 5!5!=10! = 32 ¢ 1 ¢ 2 ¢ 3 ¢ 4 ¢ 5=(6 ¢ 7 ¢ 8 ¢ 9 ¢ 10) = 8=63 =12.7 percent.
2.13 a The outcomes are pairs (x; y).
The outcome (a; a) has probability 0 to
occur. The outcome (a; b) has probability
1=4 £ 1=3 = 1=12 to occur.
The table becomes:
a b c d
a 0 1
12
1
12
1
12
b 1
12 0 1
12
1
12
c 1
12
1
12 0 1
12
d 1
12
1
12
1
12 0
2.13 b Let C be the event “c is one of the chosen possibilities”. Then C =
f(c; a); (c; b); (a; c); (b; c)g. Hence P(C) = 4=12 = 1=3.
2.14 a Since door a is never opened, P((a; a)) = P((b; a)) = P((c; a)) = 0. If the candidate
chooses a (which happens with probability 1/3), then the quizmaster chooses
without preference from doors b and c. This yields that P((a; b)) = P((a; c)) = 1=6.
If the candidate chooses b (which happens with probability 1/3), then the quizmaster
can only open door c. Hence P((b; c)) = 1=3. Similarly, P((c; b)) = 1=3. Clearly,
P((b; b)) = P((c; c)) = 0.
2.14 b If the candidate chooses a then she or he wins; hence the corresponding
event is f(a; a); (a; b); (a; c)g, and its probability is 1/3.
2.14 c To end with a the candidate should have chosen b or c. So the event is
f(b; c); (c; b)g and P(f(b; c); (c; b)g) = 2=3.
2.15 The rule is:
P(A [ B [ C) = P(A)+P(B)+P(C)¡P(A \ B)¡P(A \ C)¡P(B \ C)+P(A \ B \ C) :
That this is true can be shown by applying the sum rule twice (and using the set
property (A [ B) \ C = (A \ C) [ (B \ C)):
P(A [ B [ C) = P((A [ B) [ C) = P(A [ B) + P(C) ¡ P((A [ B) \ C)
= P(A) + P(B) ¡ P(A \ B) + P(C) ¡ P((A \ C) [ (B \ C))
= s ¡ P(A \ B) ¡ P((A \ C)) ¡ P((B \ C)) + P((A \ C) \ (B \ C))
= s ¡ P(A \ B) ¡ P(A \ C) ¡ P(B \ C) + P(A \ B \ C) :
Here we did put s := P(A) + P(B) + P(C) for typographical convenience.
2.16 Since E \ F \ G = ;, the three sets E \ F, F \ G, and E \ G are disjoint.
Since each has probability 1=3, they have probability 1 together. From these two
facts one deduces P(E) = P(E \ F)+P(E \ G) = 2=3 (make a diagram or use that
E = E \ (E \ F) [ E \ (F \ G) [ E \ (E \ G)).
2.17 Since there are two queues we use pairs (i; j) of natural numbers to indicate
the number of customers i in the first queue, and the number j in the second queue.
Since we have no reasonable bound on the number of people that will queue, we
take ­ = f(i; j) : i = 0; 1; 2; : : : ; j = 0; 1; 2; : : : g:
2.18 The probability r of no success at a certain day is equal to the probability
that both experiments fail, hence r = (1 ¡ p)2. The probability of success for the
first time on day n therefore equals rn¡1(1 ¡ r): (Cf. Section2.5.)
460 Full solutions from MIPS: DO NOT DISTRIBUTE
2.19 a We need at least two days to see two heads, hence ­ = f2; 3; 4; : : : g:
2.19 b It takes 5 tosses if and only if the fifth toss is heads (which has probability
p), and exactly one of the first 4 tosses is heads (which has probability 4p(1 ¡ p)3).
Hence the probability asked for equals 4p2(1 ¡ p)3:
3.1 Define the following events: B is the event “point B is reached on the second
step,” C is the event “the path to C is chosen on the first step,” and similarly we
define D and E. Note that the events C, D, and E are mutually exclusive and that
one of them must occur. Furthermore, that we can only reach B by first going to C
or D. For the computation we use the law of total probability, by conditioning on
the result of the first step:
P(B) = P(B \ C) + P(B \ D) + P(B \ E)
= P(B jC) P(C) + P(B jD) P(D) + P(B jE) P(E)
=
1
3
¢
1
3
+
1
4
¢
1
3
+
1
3
¢ 0 =
7
36
:
3.2 a Event A has three outcomes, event B has 11 outcomes, and A \ B =
f(1; 3); (3; 1)g. Hence we find P(B) = 11=36 and P(A \ B) = 2=36 so that
P(AjB) =
P(A \ B)
P(B)
=
2=36
11=36
=
2
11
:
3.2 b Because P(A) = 3=36 = 1=12 and this is not equal to 2=11 = P(AjB) the
events A and B are dependent.
3.3 a There are 13 spades in the deck and each has probability 1=52 of being chosen,
hence P(S1) = 13=52 = 1=4. Given that the first card is a spade there are 13¡1 = 12
spades left in the deck with 52 ¡ 1 = 51 remaining cards, so P(S2 j S1) = 12=51. If
the first card is not a spade there are 13 spades left in the deck of 51, so P(S2 j Sc
1) =
13=51.
3.3 b We use the law of total probability (based on ­ = S1 [ Sc
1):
P(S2) = P(S2 \ S1) + P(S2 \ Sc
1) = P(S2 j S1) P(S1) + P(S2 j Sc
1) P(Sc
1)
=
12
51
¢
1
4
+
13
51
¢
3
4
=
12 + 39
51 ¢ 4
=
1
4
:
3.4 We repeat the calculations from Section 3.3 based on P(B) = 1:3 ¢ 10¡5:
P(T \ B) = P(T jB) ¢ P(B) = 0:7 ¢ 0:000 013 = 0:000 0091
P(T \ Bc) = P(T jBc) ¢ P(Bc) = 0:1 ¢ 0:999 987 = 0:099 9987
so P(T) = P(T \ B) + P(T \ Bc) = 0:000 0091 + 0:099 9987 = 0:100 0078 and
P(B j T) =
P(T \ B)
P(T)
=
0:000 0091
0:100 0078
= 0:000 0910 = 9:1 ¢ 10¡5:
Further, we find
P(Tc \ B) = P(Tc jB) ¢ P(B) = 0:3 ¢ 0:000 013 = 0:000 0039
and combining this with P(Tc) = 1 ¡ P(T) = 0:899 9922:
P(B j Tc) =
P(Tc \ B)
P(Tc)
=
0:000 0039
0:899 9922
= 0:000 0043 = 4:3 ¢ 10¡6:
29.1 Full solutions 461
3.5 Define the events R1 and R2 meaning: a red ball is drawn on the first and
second draw, respectively. We are asked to compute P(R1 \ R2). By conditioning
on R1 we find:
P(R1 \ R2) = P(R2 jR1) ¢ P(R1) =
3
4
¢
1
2
=
3
8
;
where the conditional probability P(R2 jR1) follows from the contents of the urn
after R1 has occurred: one white and three red balls.
3.6 a Let E denote the event “outcome is an even numbered month” and H the
event “outcome is in the first half of the year.” Then P(E) = 1=2 and, because
in the first half of the year there are three even and three odd numbered months,
P(E jH) = 1=2 as well; the events are independent.
3.6 b Let S denote the event “outcome is a summer month”. Of the three summer
months, June and August are even numbered, so P(E j S) = 2=3 6= 1=2. Therefore,
E and S are dependent.
3.7 a The best approach to a problem like this one is to write out the conditional
probability and then see if we can somehow combine this with P(A) = 1=3 to
solve the puzzle. Note that P(B \ Ac) = P(B jAc) P(Ac) and that P(A [ B) =
P(A) + P(B \ Ac). So
P(A [ B) =
1
3
+
1
4
¢ 1 ¡
1
3=
1
3
+
1
6
=
1
2
:
3.7 b From the conditional probability we find P(Ac \ Bc) = P(Ac jBc) P(Bc) =
1
2 (1 ¡ P(B)). Recalling DeMorgan’s law we know P(Ac \ Bc) = P((A [ B)c) =
1¡P(A [ B) = 1=3. Combined this yields an equation for P(B): 1
2 (1 ¡ P(B)) = 1=3
from which we find P(B) = 1=3.
3.8 a This asks for P(W). We use the law of total probability, decomposing ­ =
F [ Fc. Note that P(W j F) = 0:99.
P(W) = P(W \ F) + P(W \ Fc) = P(W j F) P(F) + P(W j Fc) P(Fc)
= 0:99 ¢ 0:1 + 0:02 ¢ 0:9 = 0:099 + 0:018 = 0:117:
3.8 b We need to determine P(F jW), and this can be done using Bayes’ rule. Some
of the necessary computations have already been done in a, we can copy P(W \ F)
and P(W) and get:
P(F jW) =
P(F \W)
P(W)
=
0:099
0:117
= 0:846:
3.9 Deciphering the symbols we conclude that P(B jA) is to be determined. From
the probabilities listed we find P(A \ B) = P(A)+P(B)¡P(A [ B) = 3=4+2=5¡
4=5 = 7=20, so that P(B jA) = P(A \ B) =P(A) = (7=20)=(3=4) = 28=60 = 7=15.
3.10 Let K denote the event “the student knows the answer” and C the event
“the answer that is given is the correct one.” We are to determine P(K jC). From
the information given, we know that P(C jK) = 1 and P(C jKc) = 1=4, and that
P(K) = 0:6. Therefore:
P(C) = P(C jK) ¢ P(K) + P(C jKc) ¢ P(Kc) = 1 ¢ 0:6 +
1
4
¢ 0:4 = 0:6 + 0:1 = 0:7
and P(K jC) = 0:6=0:7 = 6=7:
462 Full solutions from MIPS: DO NOT DISTRIBUTE
3.11 a The probability that a driver that is classified as exceeding the legal limit in
fact does not exceed the limit.
3.11 b It is given that P(B) = 0:05. We determine the answer via P(Bc jA) =
P(Bc \ A) =P(A). We find P(Bc \ A) = P(AjBc)¢P(Bc) = 0:95 (1¡p), P(B \ A) =
P(AjB) ¢ P(B) = 0:05 p, and by adding them P(A) = 0:95 ¡ 0:9 p. So
P(Bc jA) =
0:95 (1 ¡ p)
0:95 ¡ 0:9 p
=
95 (1 ¡ p)
95 ¡ 90 p
= 0:5 when p = 0:95.
3.11 c From b we find
P(B jA) = 1 ¡ P(Bc jA) =
95 ¡ 90 p ¡ 95 (1 ¡ p)
95 ¡ 90 p
=
5p
95 ¡ 90 p
:
Setting this equal to 0:9 and solving for p yields p = 171=172 ¼ 0:9942.
3.12 We start by deriving some unconditional probabilities from what is given:
P(B \ C) = P(B jC) ¢ P(C) = 1=6 and P(A \ B \ C) = P(B \ C) ¢ P(AjB \ C) =
1=24. Next, we should realize that B\C is the union of the disjoint events A\B\C
and Ac \ B \ C, so that
P(Ac \ B \ C) = P(B \ C) ¡ P(A \ B \ C) =
1
6
¡
1
24
=
1
8
:
3.13 a There are several ways to see that P(D1) = 5=9. Method 1: the first team
we draw is irrelevant, for the second team there are always 5 “good” choices out of 9
remaining teams. Method 2: there are 􀀀10
2 = 45 possible outcomes when two teams
are drawn; among these, there are 5 ¢ 5 = 25 favorable outcomes (all weak-strong
pairings), resulting in P(D1) = 25=45 = 5=9.
3.13 b Given D1, there are 4 strong and 4 weak teams left. Using one of the methods
from a on this reduced number of teams, we find P(D2 jD1) = 4=7 and P(D1 \ D2) =
P(D2 jD1) ¢ P(D1) = (4=7) ¢ (5=9) = 20=63.
3.13 c Proceeding in the same fashion, we find P(D3 jD1 \ D2) = 3=5 and
P(D1 \ D2 \ D3) = P(D3 jD1 \ D2) ¢ P(D1 \ D2) = (3=5) ¢ (20=63) = 12=63.
3.13 d Subsequently, we find P(D4 jD1\¢ ¢ ¢\D3) = 2=3 and P(D5 jD1\¢ ¢ ¢\D4) =
1. The final result can be written as
P(D1 \ ¢ ¢ ¢ \ D5) =
5
9
¢
4
7
¢
3
5
¢
2
3
¢ 1 =
8
63
:
The probability of a “dream draw” with n strong and n weak teams is P(D1 \ ¢ ¢ ¢ \ Dn) =
2n=􀀀2n
n .
3.14 a If you chose the right door, switching will make you lose, so P(W jR) = 0.
If you chose the wrong door, switching will make you win for sure: P(W jRc) = 1.
3.14 b Using P(R) = 1=3, we find:
P(W) = P(W \ R) + P(W \ Rc) = P(W jR) P(R) + P(W jRc) P(Rc)
= 0 ¢
1
3
+ 1 ¢
2
3
=
2
3
:
29.1 Full solutions 463
3.15 This is a puzzle and there are many ways to solve it. First note that P(A) =
1=2. We condition on A [ B:
P(B) = P(B jA [ B) ¢ P(A [ B)
=
2
3
fP(A) + P(B) ¡ P(A) P(B)g
=
2
3
f
1
2
+ P(B) ¡
1
2
P(B)g
=
1
3
+
1
3
P(B) :
Solving the resulting equation yields P(B) =
13
1¡13
= 1
2 .
3.16 a Using Bayes’ rule we find:
P(Dj T) =
P(D \ T)
P(T)
=
0:98 ¢ 0:01
0:98 ¢ 0:01 + 0:05 ¢ 0:99
¼ 0:1653:
3.16 b Denote the event “the second test indicates you have the disease” with the
letter S.
Method 1 (“unconscious”): from the first test we know that the probability we have
the disease is not 0:01 but 0:1653 and we reason that we shoud redo the calculation
from a with P(D) = 0:1653:
P(Dj S) =
P(D \ S)
P(S)
=
0:98 ¢ 0:1653
0:98 ¢ 0:1653 + 0:05 ¢ 0:8347
¼ 0:7951:
This is the correct answer, but a more thorough consideration is warrented.
Method 2 (“conscientious”): we are in fact to determine
P(Dj S \ T) =
P(D \ S \ T)
P(S \ T)
and we should wonder what “independent repetition of the test” exactly means.
Clearly, both tests are dependent on whether you have the disease or not (and as a
result, S and T are dependent), but given that you have the disease the outcomes
are independent, and the same when you do not have the disease. Formally put:
given D, the events S and T are independent; given Dc, the events S and T are
independent. In formulae:
P(S \ T jD) = P(S jD) ¢ P(T jD);
P(S \ T jDc) = P(S jDc) ¢ P(T jDc):
So P(D \ S \ T) = P(S jD) ¢ P(T jD)P(D) = (0:98)2 ¢ 0:01 = 0:009604 and
P(Dc \ S \ T) = (0:05)2 ¢0:99 = 0:002475. Adding them yields P(S \ T) = 0:012079
and so P(Dj S \ T) = 0:009604=0:012079 ¼ 0:7951. Note that P(S j T) ¼ 0:2037
which is much larger than P(S) ¼ 0:0593 (and for a good test, it should be).
3.17 a I win the game after the next two rallies if I win both: P(W \ G) = p2.
Similarly for losing the game if you win both rallies: P(Wc \ G) = (1 ¡ p)2. So
P(W jG) = p2=(p2 + (1 ¡ p)2).