Page 31 - Probability and Statistical Inference

8 1. Nations of Probability

(ii) Observe that A ∪ A = S and then proceed as before. Observe that A
c
and A are disjoint events. "
c
(iii) Notice that B = (B ∩ A) ∪ (B ∩ A ) where B ∩ A and B ∩ A are
c
c
disjoint events. Hence by part (iii) in the Definition 1.3.4, we claim that

Now, the result is immediate. "
c
(iv) It is easy to verify that A ∪ B = (A ∩ B ) ∪ (B ∩ A ) ∪ (A ∩ B)
c
c
c
where the three events A ∩ B , B ∩ A , A ∩ B are also disjoint. Thus, we
have

which leads to the desired result. "
(v) We leave out its proof as the Exercise 1.3.4. "
(vi) Since the sequence of events {B ; i ≥ 1} forms a partition of the
sample space S, we can write i
where the events A ∩ B , i = 1,2, ... are also disjoint. Now, the result follows
i
from part (iii) in the Definition 1.3.4. !
Example 1.3.1 (Example 1.1.1 Continued) Let us define three events as
follows:

How can we obtain the probabilities of these events? First notice that as sub-
sets of S, we can rewrite these events as A = {HHT, HTH, THH}, B = {HHT,
HTH, HTT, THH, THT, TTH, TTT}, and C = {TTT}. Now it becomes obvious
that P(A) = 3/8, P(B) = 7/8, and P(C) = 1/8. One can also see that A ∩ B =
{HHT, HTH, THH} so that P(A ∩ B) = 3/8, whereas A ∪ C = {HHT, HTH,
THH, TTT} so that P(A ∪ C) = 4/8 = 1/2.
Example 1.3.2 Example 1.1.2 Continued) Consider the following events:

Now, as subsets of the corresponding sample space S, we can rewrite these
events as D = {26, 35, 44, 53, 62} and E = {31, 42, 53, 64}. It is now obvious
that P(D) = 5/36 and P(E) = 4/36 = 1/9. !