Page 478 - Discrete Mathematics and Its Applications

7.2 Probability Theory 457

Because E ∩ F ={0000, 0001, 0010, 0011, 0100}, we see that p(E ∩ F) = 5/16. Because
there are eight bit strings of length four that start with a 0, we have p(F) = 8/16 = 1/2.
Consequently,
5/16 5 ▲
p(E | F) = = .
1/2 8
EXAMPLE 4 What is the conditional probability that a family with two children has two boys, given they
have at least one boy? Assume that each of the possibilities BB, BG, GB, and GG is equally
likely, where B represents a boy and G represents a girl. (Note that BG represents a family with
an older boy and a younger girl while GB represents a family with an older girl and a younger
boy.)

Solution: Let E be the event that a family with two children has two boys, and let F be
the event that a family with two children has at least one boy. It follows that E ={BB},
F ={BB, BG, GB}, and E ∩ F ={BB}. Because the four possibilities are equally likely, it
follows that p(F) = 3/4 and p(E ∩ F) = 1/4. We conclude that
p(E ∩ F) 1/4 1 ▲
p(E | F) = = = .
p(F) 3/4 3

Independence

Suppose a coin is flipped three times, as described in the introduction to our discussion of
conditional probability. Does knowing that the first flip comes up tails (event F) alter the
probability that tails comes up an odd number of times (event E)? In other words, is it the case
that p(E | F) = p(E)? This equality is valid for the events E and F, because p(E | F) = 1/2
and p(E) = 1/2. Because this equality holds, we say that E and F are independent events.
When two events are independent, the occurrence of one of the events gives no information
about the probability that the other event occurs.
Because p(E | F) = p(E ∩ F)/p(F), asking whether p(E | F) = p(E) is the same as
asking whether p(E ∩ F) = p(E)p(F). This leads to Definition 4.

DEFINITION 4 The events E and F are independent if and only if p(E ∩ F) = p(E)p(F).

EXAMPLE 5 Suppose E is the event that a randomly generated bit string of length four begins with a 1
and F is the event that this bit string contains an even number of 1s. Are E and F independent,
if the 16 bit strings of length four are equally likely?

Solution: There are eight bit strings of length four that begin with a one: 1000, 1001, 1010,
1011, 1100, 1101, 1110, and 1111. There are also eight bit strings of length four that contain
an even number of ones: 0000, 0011, 0101, 0110, 1001, 1010, 1100, 1111. Because there are
16 bit strings of length four, it follows that

p(E) = p(F) = 8/16 = 1/2.
Because E ∩ F ={1111, 1100, 1010, 1001}, we see that

p(E ∩ F) = 4/16 = 1/4.
Because

p(E ∩ F) = 1/4 = (1/2)(1/2) = p(E)p(F),
we conclude that E and F are independent. ▲