563489578934

684
which is identical to Eq. (B–4), since
Probability and Random Variables
Appendix B
P(A) = lim an 1 + n AB
b,
n: q n
P(B) = lim an 2 + n AB
b, and P(AB) = lim
n: q n
n b
n:q an AB
Example B–1 (Continued)
The probability of having a blocked intersection or rain occurring or both is then
P(A + B) = 0.0025 + 0.03 - 0.002 L 0.03
(B–5)
Conditional Probabilities
DEFINITION. The probability that an event A occurs, given that an event B has also
occurred, is denoted by P(A|B), which is defined as
Example B–1 (Continued)
P(A|B) =
lim an AB
b
n B : q n B
The probability of a blocked intersection when it is raining is approximately
P(A|B) = 20
300 = 0.066
(B–6)
(B–7)
THEOREM.
This is known as Bayes’ theorem.
Proof.
Let E = AB; then
P(AB) = P(A)P(B ƒ A) = P(B)P(A ƒ B)
(B–8)
P(AB) = lim
n: q an AB
n b =
(B–9)
It is noted that the values obtained for P(AB), P(B), and P(A|B) in Example B–1 can be verified
by using Eq. (B–8).
DEFINITION.
lim
n: q
n A :large
a n AB
n A
Two events, A and B, are said to be independent if either
P(A|B) = P(A)
n A
b = P(B ƒ A)P(A)
n
(B–10)
or
P(B|A) = P(B)
(B–11)
Using this definition, we can easily demonstrate that events A and B of Example B–1 are
not independent. Conversely, if event A had been defined as getting heads on a coin toss,