Introduction to Probability and Statistics for Engineers and Scientists

70 Chapter 3:Elements of Probability

the dinner is equally likely to be a boy or a girl. Their mistake, however, is in assuming that
these two possibilities are equally likely. Remember that initially there were four equally
likely outcomes. Now the information that at least one child is a boy is equivalent to
knowing that the outcome is not (g,g). Hence we are left with the three equally likely
outcomes (b,b), (b,g), (g,b), thus showing that the Jones child not attending the dinner
is twice as likely to be a girl as a boy. ■

By multiplying both sides of Equation 3.6.1 byP(F) we obtain that

P(EF)=P(F)P(E|F) (3.6.2)

In words, Equation 3.6.2 states that the probability that bothEandFoccur is equal to
the probability thatFoccurs multiplied by the conditional probability ofEgiven that
Foccurred. Equation 3.6.2 is often quite useful in computing the probability of the
intersection of events. This is illustrated by the following example.

EXAMPLE 3.6c Ms. Perez figures that there is a 30 percent chance that her company will
set up a branch office in Phoenix. If it does, she is 60 percent certain that she will be
made manager of this new operation. What is the probability that Perez will be a Phoenix
branch office manager?

SOLUTION If we letBdenote the event that the company sets up a branch office in Phoenix
andMthe event that Perez is made the Phoenix manager, then the desired probability is
P(BM), which is obtained as follows:

P(BM)=P(B)P(M|B)

=(.3)(.6)

=.18

Hence, there is an 18 percent chance that Perez will be the Phoenix manager. ■

3.7Bayes’ Formula

LetEandFbe events. We may expressEas

E=EF∪EFc

for, in order for a point to be inE, it must either be in bothEandFor be inEbut not inF.
(See Figure 3.6.) AsEFandEFcare clearly mutually exclusive, we have by Axiom 3 that

P(E)=P(EF)+P(EFc)

=P(E|F)P(F)+P(E|Fc)P(Fc)

=P(E|F)P(F)+P(E|Fc)[ 1 −P(F)] (3.7.1)