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

142 PROBABILITY [CHAP. 7

(d) We have

P(neitherAnorB)=P ((A∪B)C)= 1 −P(A∪B)= 1 − 157 = 158

7.20. Consider the following events for a family with children:

A=(children of both sexes},B={at most one boy}.

(a) Show thatAandBare independent events if a family has three children.

(b) Show thatAandBare dependent events if a family has only two children.

(a) We have the equiprobable spaceS={bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}. Here

A={bbg, bgb, bgg, gbb, gbg, ggb} and so P (A)=^68 =^34

B={bgg, gbg, ggb, ggg} and so P(B)=^48 =^12

A∩B={bgg, gbg, ggb} and so P(A∩B)=^38

SinceP (A)P (B)=^34 ·^12 =^38 =P(A∩B) AandBare independent.

(b) We have the equiprobable spaceS={bb, bg, gb, gg}. Here

A={bg, gb} and so P (A)=^12

B={bg, gb, gg} and so P(B)=^34

A∩B={bg, gb} and so P(A∩B)=^12

SinceP (A)P (B)=P(A∩B),AandBare dependent.

7.21. BoxAcontains five red marbles and three blue marbles, and boxBcontains three red and two blue.

A marble is drawn at random from each box.

(a) Find the probabilitypthat both marbles are red.

(b) Find the probabilitypthat one is red and one is blue.

(a) The probability of choosing a red marble fromAis^58 and fromBis^35. Since the events are independent,

P=^58 ·^35 =^38.

(b) The probabilityp 1 of choosing a red marble fromAand a blue marble fromBis^58 ·^25 =^14. The probabilityp 2

of choosing a blue marble fromAand a red marble fromBis^38 ·^35 = 409. Hencep=p 1 +p 2 =^14 + 409 =^1940.

7.22. Prove: IfAandBare independent events, thenAcandBcare independent events.

LetP (A)=xandP(B)=y. ThenP(Ac)= 1 −xandP(Bc)= 1 −y. SinceAandBare independent.

P(A∩B)=P (A)P (B)=xy. Furthermore,

P(A∪B)=P (A)+P(B)−P(A∩B)=x+y−xy

By DeMorgan’s law,(A∪B)c=Ac∩Bc; hence

P(Ac∩Bc)=P ((A∪B)c)= 1 −P(A∪B)= 1 −x−y+xy

On the other hand,

P(Ac)P (Bc)=( 1 −x)( 1 −y)= 1 −x−y+xy

ThusP(Ac∩Bc)=P(Ac)P (Bc), and soAcandBcare independent.

In similar fashion, we can show thatAandBc, as well asAcandB, are independent.