Introduction to Probability and Statistics for Engineers and Scientists

3.8Independent Events 77

words, knowing thatFhas occurred generally changes the chances ofE’s occurrence. In
the special cases whereP(E|F) does in fact equalP(E), we say thatEis independent
ofF. That is,Eis independent ofFif knowledge thatFhas occurred does not change
the probability thatEoccurs.
SinceP(E|F)=P(EF)/P(F), we see thatEis independent ofFif

P(EF)=P(E)P(F) (3.8.1)

Since this equation is symmetric inEandF, it shows that wheneverEis independent of
Fso isFofE. We thus have the following.

Definition

Two eventsEandFare said to beindependentif Equation 3.8.1 holds. Two eventsE
andFthat are not independent are said to bedependent.

EXAMPLE 3.8a A card is selected at random from an ordinary deck of 52 playing cards. If
Ais the event that the selected card is an ace andHis the event that it is a heart, thenA
andHare independent, sinceP(AH)= 521 , whileP(A)= 524 andP(H)=^1352. ■

EXAMPLE 3.8b If we letEdenote the event that the next president is a Republican and
Fthe event that there will be a major earthquake within the next year, then most people
would probably be willing to assume thatEandFare independent. However, there would
probably be some controversy over whether it is reasonable to assume thatEis independent
ofG, whereGis the event that there will be a recession within the next two years. ■

We now show that ifEis independent ofFthenEis also independent ofFc.

PROPOSITION 3.8.1 IfEandFare independent, then so areEandFc.

Proof

Assume thatEandFare independent. SinceE=EF∪EFc, andEFandEFcare obvi-
ously mutually exclusive, we have that

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

=P(E)P(F)+P(EFc) by the independence ofEandF

or equivalently,

P(EFc)=P(E)(1−P(F))

=P(E)P(Fc)

and the result is proven.