Page 477 - Discrete Mathematics and Its Applications

456 7 / Discrete Probability

Also, note that if the events E 1 and E 2 are disjoint, then p(E 1 ∩ E 2 ) = 0, which implies
that

p(E 1 ∪ E 2 ) = p(E 1 ) + p(E 2 ) − p(E 1 ∩ E 2 ) = p(E 1 ) + p(E 2 ).

Theorem 1 generalizes this last formula by providing a formula for the probability of the
union of pairwise disjoint events.

THEOREM 1 If E 1 ,E 2 ,... is a sequence of pairwise disjoint events in a sample space S, then

p E i = p(E i ).
i i
(Note that this theorem applies when the sequence E 1 ,E 2 ,... consists of a finite number or
a countably infinite number of pairwise disjoint events.)

We leave the proof of Theorem 1 to the reader (see Exercises 36 and 37).

Conditional Probability

Suppose that we flip a coin three times, and all eight possibilities are equally likely. Moreover,
supposeweknowthattheeventF,thatthefirstflipcomesuptails,occurs.Giventhisinformation,
what is the probability of the event E, that an odd number of tails appears? Because the first
flip comes up tails, there are only four possible outcomes: TTT, TTH, THT, and THH, where H
and T represent heads and tails, respectively. An odd number of tails appears only for the
outcomes TTT and THH. Because the eight outcomes have equal probability, each of the four
possible outcomes, given that F occurs, should also have an equal probability of 1/4. This
suggests that we should assign the probability of 2/4 = 1/2to E, given that F occurs. This
probability is called the conditional probability of E given F.
In general, to find the conditional probability of E given F,weuse F as the sample space.
For an outcome from E to occur, this outcome must also belong to E ∩ F. With this motivation,
we make Definition 3.

DEFINITION 3 Let E and F be events with p(F) > 0. The conditional probability of E given F, denoted
by p(E | F), is defined as

p(E ∩ F)
p(E | F) = .
p(F)

EXAMPLE 3 A bit string of length four is generated at random so that each of the 16 bit strings of length four
is equally likely. What is the probability that it contains at least two consecutive 0s, given that
its first bit is a 0? (We assume that 0 bits and 1 bits are equally likely.)
Solution: Let E be the event that a bit string of length four contains at least two consecutive 0s,
and let F be the event that the first bit of a bit string of length four is a 0. The probability that a
bit string of length four has at least two consecutive 0s, given that its first bit is a 0, equals

p(E ∩ F)
p(E | F) = .
p(F)