Begin2.DVI

of the probabilities associated with the simple events common to both the eventsE 1

andE 2 are counted twice. The sum of these common probabilities of simple events

which are counted twice isP(E 1 ∩E 2 )and so this value is subtracted from the sum

P(E 1 ) +P(E 2 ).

Example 11-1.

Two coins are tossed. What is the probability that at least one tail occurs?

Assume the coins are not trick coins so that there are four equally likely events that

can occur. LetH denote heads andT denote tails, then the sample space for the

experiment isS={HH, HT, T H, T T}. Because each event is equally likely, assign a

probability of 1 / 4 to each event. For this example, the eventEto be investigated is

E={HT}∪{T H}∪{T T}

That is, at least one tail occurs. Consequently,

P(E) =P(HT) +P(T H) +P(T T) = 1/4 + 1/4 + 1/4 = 3/ 4

Example 11-2.

A pair of fair dice are rolled. The sample space associated with this experiment

is a representation of all possible outcomes.

S={(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)

(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)

(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)

(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)

(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)}

There are 36 equally likely possible outcomes. Assign a probability of 1 / 36 to each

simple event.

IfE 1 is the event that a 7 is rolled, then

P(E 1 ) =P((1,6)) +P((2,5)) +P((3,4)) +P((4,3)) +P((5,2)) +P((6,1)) = 6/36 = 1/ 6