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

CHAP. 7] PROBABILITY 137

(a) Express explicitly the following events:

A={heads and an even number}, B={prime number}, C={tails and an odd number}

(b) Express explicitly the events: (i)AorBoccurs; (ii)BandCoccur: (iii) onlyBoccurs.

(c) Which pair of the eventsA,B, andCare mutually exclusive?

(a) The elements ofAare those elements ofSconsisting of anHand an even number:

A={H 2 ,H 4 ,H 6 }

The elements ofBare those points inSwhose second component is a prime number (2, 3, or 5):

B={H 2 ,H 3 ,H 5 ,T 2 ,T 3 ,T 5 }

The elements ofCare those points inSconsisting of aTand an odd number;C={T 1 ,T 3 ,T 5 }.

(b) (i) A∪B={H 2 ,H 4 ,H 6 ,H 3 ,H 5 ,T 2 ,T 3 ,T 5 }

(ii) B∩C={T 3 ,T 5 }

(iii) B∩Ac∩Cc={H 3 ,H 5 ,T 2 }

(c) AandCare mutually exclusive sinceA∩C=
.

7.2.A pair of dice is tossed. (See Example 7.2.) Find the number of elements in each event:

(a)A={two numbers are equal} (c)C={5 appears on the first die}

(b)B={sum is 10 or more} (d)D={5 appears on at least one die}

Use Fig. 7-1(b)to help count the number of elements in the event.

(a) A={( 1 , 1 ), ( 2 , 2 ),...,( 6 , 6 )},son(A)= 6.

(b) B={( 6 , 4 ), ( 5 , 5 ), ( 4 , 6 ), ( 6 , 5 ), ( 5 , 6 ), ( 6 , 6 )},son(B)=6.

(c) C={( 5 , 1 ), ( 5 , 2 ),...,( 5 , 6 )},son(C)=6.

(d) There are six pairs with 5 as the first element, and six pairs with 5 as the second element. However, (5, 5) appears

in both places. Hence

n(D)= 6 + 6 − 1 = 11

Alternately, count the pairs in Fig. 7-1(b)which are inDto getn(D)=11.

FINITE EQUIPROBABLE SPACES

7.3 Determine the probabilitypof each event:

(a) An even number appears in the toss of a fair die;

(b) One or more heads appear in the toss of three fair coins;

(c) A red marble appears in a random drawing of one marble from a box containing four white, three red,

and five blue marbles.

Each sample spaceSis an equiprobable space. Thus, for each eventE, use:

P(E)=

number of elements inE

number of elements inS

=

n(E)

n(S)

(a) The event can occur in three ways (2, 4 or 6) out of 6 cases; hencep=^36 =^12.

(b) There are 8 cases:

HHH,HHT,HTH,HTT,THH,THT,TTH,TTT

Only the last case is not favorable; hencep= 7 /8.

(c) There are 4+ 3 + 5 =12 marbles of which three are red; hencep= 123 =^14.