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

130 PROBABILITY [CHAP. 7

Also,

P(A∩B)=P({HHH,HHT})=^14 ,P(A∩C)=P({HHT})=^18 ,P(B∩C)=P({HHT,THH})=^14

Accordingly,

P (A)P (B) =^12 ·^12 =^14 =P(A∩B), and soAandBare independent

P (A)P (C) =^12 ·^14 =^18 =P(A∩C), and soAandCare independent

P(B)P(C) =^12 ·^14 =^18 =P(B∩C), and soBandCare dependent

Frequently, we will postulate that two events are independent, or the experiment itself will imply that
two events are independent.

EXAMPLE 7.10 The probability thatAhits a target is^14 , and the probability thatBhits the target is^25. Both
shoot at the target. Find the probability that at least one of them hits the target, i.e., thatAorB(or both) hit the
target.
We are given thatP (A)=^12 andP(B)=^25 , and we seekP(A∪B). Furthermore, the probability thatAor
Bhits the target is not influenced by what the other does; that is, the event thatAhits the target is independent
of the event thatBhits the target, that is,P(A∩B)=P (A)P (B). Thus

P(A∪B)=P (A)+P(B)−P(A∩B)=P (A)+P(B)−P (A)P (B)=^14 +^25 −

(

1

4

)(

2

5

)

=^1120

7.6Independent Repeated Trials, Binomial Distribution

We have previously discussed probability spaces which were associated with an experiment repeated a finite
number of times, as the tossing of a coin three times. This concept of repetition is formalized as follows:

Definition 7.3: LetSbe a finite probability space. By the space ofn independent repeated trials, we mean the
probability spaceSnconsisting of orderedn-tuples of elements ofS, with the probability of ann-tuple defined to
be the product of the probabilities of its components:

P ((s 1 ,s 2 ,...,sn))=P(s 1 )P (s 2 )...P(sn)

EXAMPLE 7.11 Whenever three horsesa,b, andcrace together, their respective probabilities of winning are
1
2 ,

1

3 , and

1

6. In other words,S={a, b, c}withP(a)=

1

2 ,P(b)=

1

3 , andP(c)=

1
6. If the horses race twice,
then the sample space of the two repeated trials is

S 2 ={aa, ab, ac, ba, bb, bc, ca, cb, cc}

For notational convenience, we have writtenacfor the ordered pair(a, c). The probability of each point inS 2 is

P (aa)=P(a)P(a)=^12

(

1

2

)

=^14 ,P(ba)=^16 ,P(ca)= 121

P(ab)=P(a)P(b)=^12

(

1

3

)

=^16 , P (bb)=^19 , P (cb)= 181

P(ac)=P(a)P(c)=^12

(

1

6

)

= 121 , P (bc)= 181 , P (cc)= 361

Thus the probability ofcwinning the first race andawinning the second race isP(ca)= 121.