Introduction to Probability and Statistics for Engineers and Scientists

3.5Sample Spaces Having Equally Likely Outcomes 61

S

EF

I II III

FIGURE 3.4

which shows that

P(E∪F)=P(E)+P(F)−P(II)

and the proof is complete since II=EF.

EXAMPLE 3.4a A total of 28 percent of American males smoke cigarettes, 7 percent smoke
cigars, and 5 percent smoke both cigars and cigarettes. What percentage of males smoke
neither cigars nor cigarettes?

SOLUTION LetEbe the event that a randomly chosen male is a cigarette smoker and letB
be the event that he is a cigar smoker. Then, the probability this person is either a cigarette
or a cigar smoker is

P(E∪F)=P(E)+P(F)−P(EF)=.07+.28−.05=.3

Thus the probability that the person is not a smoker is .7, implying that 70 percent of
American males smoke neither cigarettes nor cigars. ■

Theoddsof an eventAis defined by

P(A)

P(Ac)

=

P(A)

1 −P(A)

Thus the odds of an eventAtells how much more likely it is thatAoccurs than that it
does not occur. For instance ifP(A)=3/4, thenP(A)/(1−P(A))=3, so the odds is 3.
Consequently, it is 3 times as likely thatAoccurs as it is that it does not.

3.5Sample Spaces Having Equally Likely Outcomes

For a large number of experiments, it is natural to assume that each point in the sample
space is equally likely to occur. That is, for many experiments whose sample spaceSis a
finite set, sayS={1, 2,...,N}, it is often natural to assume that

P({ 1 })=P({ 2 })= ··· =P({N})=p (say)