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CONDITIONAL PROBABILITIES
Suppose that we toss 2 dice, and suppose that each of the 36 possible outcomes is
equally likely to occur and hence has probability 1
36 . Suppose further that we observethat the first die is a 3. Then, given this information, what is the probability that the
sum of the 2 dice equals 8? To calculate this probability, we reason as follows: Given
that the initialdie is a 3, there can be at most 6 possible outcomes of our experiment,
namely, (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), and (3, 6). Since each of these outcomes
originally had the same probability ofoccurring, the outcomes should still have equal
probabilities. That is, given that the first die is a 3, the (conditional) probability of
each of the outcomes (3, 1), (3, 2), (3, 3), (3, 4), (3, 5),and (3, 6) is 16
, whereas the
(conditional) probability of the other 30 points in the sample space is 0. Hence, the
desired probability will be 16
.
If we let E and F denote, respectively, theevent that the sum of the dice is 8 and the
event that the first die is a 3, then the probability just obtained is called the conditional
probability that E occurs given that F has occurred and isdenoted by
P(E|F)
Ageneral formula for P(E|F) that is valid for all events E and F is derived in the same
manner: If the event F occurs, then, in order for E to occur, it is necessary that theactual occurrence be a point both in E and in F; that is, it must be in EF. Now, since
we know that F has occurred, it follows that F becomes our new, or reduced, sample
space; hence, the probabilitythat the event EF occurs will equal the probability of
EF relative to the probability of F. That is, we have the following definition.

Definition
If P(F) > 0, then
P(E|F) = P(EF)P(F)

Example:

A student is taking a one-hour-time-limit makeup examination. Suppose the probability
that the student will finish the exam in less than x hours is x/2, for all 0 … x … 1.
Then,...