Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

CHAP. 7] PROBABILITY 139

FINITE PROBABILITY SPACES

7.9.A sample spaceSconsists of four elements; that is,S={a 1 ,a 2 ,a 3 ,a 4 }. Under which of the following

functions doesSbecome a probability space?

(a)P(a 1 )=^12 P(a 2 )=^13 P(a 3 )=^14 P(a 4 )=^15

(b)P(a 1 )=^12 P(a 2 )=^14 P(a 3 )=−^14 P(a 4 )=^12

(c)P(a 1 )=^12 P(a 2 )=^14 P(a 3 )=^18 P(a 4 )=^18

(d)P(a 1 )=^12 P(a 2 )=^14 P(a 3 )=^14 P(a 4 )= 0

(a) Since the sum of the values on the sample points is greater than one, the function does not defineSas a probability

space.

(b) SinceP(a 3 )is negative, the function does not defineSas a probability space.

(c) Since each value is nonnegative and the sum of the values is one, the function does defineSas a probability space.

(d) The values are nonnegative and add up to one; hence the function does defineSas a probability space.

7.10.A coin is weighted so that heads is twice as likely to appear as tails. FindP(T)andP(H).

LetP(T)=p; thenP(H)= 2 p. Now set the sum of the probabilities equal to one, that is, setp+ 2 p=1.

Thenp=^13. ThusP(H)=^13 andP(T)=^23.

7.11. SupposeAandBare events withP (A)= 0 .6,P(B)= 0 .3, andP(A∩B)= 0 .2. Find the probability

that:

(a)Adoes not occur; (c)AorBoccurs;

(b)Bdoes not occur; (d) NeitherAnorBoccurs.

(a) P(notA)=P(AC)= 1 −P (A)= 0 .4.

(b) P(notB)=P(BC)= 1 −P(B)= 0 .7.

(c) By the Addition Principle,

P(AorB) = P(A∪B)=P (A)+P(B)−P(A∩B)

= 0. 6 + 0. 3 − 0. 2 = 0. 7

(d) Recall (DeMorgan’s Law) that neitherAnorBis the complement ofA∪B. Thus:

P(neitherAnorB)=P ((A∪B)C)= 1 −P(A∪B)= 1 − 0. 7 = 0. 3

7.12. Prove Theorem 7.2:P(Ac)= 1 −P (A).

S=A∪AcwhereAandAcare disjoint. Our result follows from the following:

1 =P(S)=P(A∪Ac)=P (A)+P(Ac)

7.13. Prove Theorem 7.3: (i)P(
)=0; (ii)P(A\B)=P (A)−P(A∩B); (iii) IfA⊆B, thenP (A)≤P(B).

(i) 
=ScandP(S)=1. ThusP(
)= 1 − 1 =0.

(ii) As indicated by Fig. 7-6(a),A=(A\B)∪(A∩B)whereA\BandA∩Bare disjoint. Hence

P (A)=P(A\B)+P(A∩B)

From which our result follows.

(iii) IfA⊆B, then, as indicated by Fig. 7-6(b),B=A∪(B\A)whereAandB\Aare disjoint. Hence

P(B)=P (A)+P(B\A)

SinceP(B\A)≥0, we haveP (A)≤P(B).