TEST BANK FOR A Modern Introduction to Probability and Statistics_ Understanding Why an By Micheal

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.)
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