Section 4.6 Probability
305
Now suppose the die is loaded so that a 4 comes up three times more often
than a 1, 2, 3, 5, or 6. We can describe the set of outcomes for the loaded die by
{1, 2, 3, 4 1 , 4 2 , 4 3 , 5, 6}
The size of the sample space is now 0 S 0 = 8 and the probability of rolling a 3 is now
P(T ) =
0 T 0
0 S 0
=
1
8
= 0.125
This is a lower probability than before because the loaded die is not as likely to
come up with a 3. However, if F is the event of rolling a 4, then there are three
successful outcomes from the sample space. Therefore, the probability of rolling
a 4 is now
P(F ) =
0 F 0
0 S 0
=
3
8
= 0.375
which is higher than before because the loaded die is more likely to come up
with a 4.
Another way to look at problems where not all outcomes are equally likely is
to assign a probability distribution to the sample space. Rather than artificially
enlarging the sample space by creating duplicates of outcomes that occur more
frequently, simply consider each distinct outcome in the original sample space as
an event and assign it a probability. If there are k different outcomes in the sample space and each outcome x i is assigned a probability p(x i ), the following rules
apply:
1. 0 ≤ p(x i ) ≤ 1
2. ∙
k
i=1
p(x i ) = 1
The first equation must hold because any probability value must fall within this
range. The second equation must hold from observation 6 in Table 4.3; the union
of all of these k disjoint outcomes is the sample space S, and the probability of S
is 1.
Now consider some event E # S. The probability of event E is then given by
P(E) = ∙
x i [E
p(x i )
(1)
In other words, we can add up all the probabilities for the individual outcomes
in E. This also follows from observation 6 in Table 4.3; E is the union of all its
distinct outcomes. The definition of P(E) as 0 E 0 / 0 S 0 when the outcomes are equally
likely is a special case of this definition where p(x i ) = 1/ 0 S 0 for each x i in E.