510 CHAPTER 8 Discrete Probability
one another in the following sense. Imagine that we can buy a chance on winning a prize
if the die is rolled and C occurs-that is, if a 1, 2, 3, or 4 is rolled. We are hesitating about
buying a chance when we learn that the die has just been rolled and that B occurred (a 1,
4, or 5 turned up). We are offered a last minute opportunity to buy a chance on C-that is,
to bet that C also occurred when the die was rolled. Are we more tempted than before to
buy a chance on C?
Here is a possible line of reasoning: We have new information, so we can revise the
sample space from E2 = {1, 2, 3, 4, 5, 6) to 021 = {1, 4, 5) = B to takeinto account the fact
that B has occurred. Initially, we assumed that all outcomes in B were equally likely. Since
we have no reason to believe otherwise now, we persist in this assumption. This means that
our probability distribution on Q I is uniform, with p1(1) = pl(^4 ) = pl(^5 ) = 1/3. Now,
for C to have occurred, it must be that B n C = { 1, 4} occurred. The revised probability is
2
PI(B Cnc) = pi(1) + pi(4) = - 3
This, however, is exactly the same as our initial assessment of C (before we knew that B
occurred): The original probability P(C) was also 2/3. Therefore, the information that B
occurred does not cause us to revise our predictions about C. In this sense, B and C are
unrelated events.
On the other hand, if we had an opportunity to buy a last-minute chance on A
11, 2, 3), we would be less tempted: A occurred only if A n B occurred. The revised prob-
ability of this is
1
PI(A n B) = P 1 ({1}) = pi(l) = -
3
The initial probability of A was P (A) = 1/2.
The reasoning we just used is widely accepted and forms the basis for a standard
technique to revise probabilities in light of new information. If events A and B are
independent-that is, if they are events in the same sample space such that P(A fl B) =
P (A) • P (B)-then knowledge that one of these events occurred or did not occur does not
enable us to revise the probability for the occurrence of the other. To explain this, we must
first say what we mean by the revised probability of an event because of the occurrence of
another event. Then, we will note that the revised probability is equal to the initial prob-
ability precisely when the two events are independent. The formula we will develop for
revised probability is also useful in its own right.
To devise a formula for what we mean by the probability of an event B C 02, given
that an event A has occurred where A C Q2 and P(A) > 0, we reason as follows: Since A
has occurred, the sample space E2 with probability density function p can be revised to a
new sample space 21 = A, for surely A includes all the outcomes that are possible. Next,
we devise a probability density function pl for the outcomes in i2 1 = A. Of course, the
sum of the pl (w)'s for to E A must be one, and pI (co) must be greater than zero if pI is to
be a density function. Also, we want the ratios of probability densities for outcomes in A
to be the same for the new density as for the old. For example, if 0i and wOj are outcomes in
A such that p(wi) =^2 p(Woj), then pl (wi) should be twice pl (w j). We want this property,
since we have no grounds for believing that the frequencies of outcomes have changed
relative to one another just because A occurred. To obtain a legitimate probability density
for the new sample space Q I = A that preserves the original ratios of probabilities, we can