Page 21 - Schaum's Outlines - Probability, Random Variables And Random Processes

PROBABILITY [CHAP 1

1.12. Verify the distributive law (1.1 2).
Let s E [A n (B u C)]. Then s E A and s E (B u C). This means either that s E A and s E B or that
s E A and s E C; that is, s E (A n B) or s E (A n C). Therefore,
A n (B u C)c [(A n B) u (A n C)]
Next, let s E [(A n B) u (A n C)]. Then s E A and s E B or s E A and s E C. Thus s E A and (s E B or
s E C). Thus,
[(A n B) u (A n C)] c A n (B u C)
Thus, by the definition of equality, we have
A n (B u C)= (A n B) u (A n C)

1.13. Using a Venn diagram, repeat Prob. 1.12.
Figure 1-6 shows the sequence of relevant Venn diagrams. Comparing Fig. 1-6(b) and 1-6(e), we con-
clude that

(u) Shaded region: H u C' (h) Shaded region: A n (B u C)

(c) Shaded region: A n H ((1) Shaded region: A n C

(r) Shaded region: (A n H) u (A n C)
Fig. 1-6

1.14. Let A and B be arbitrary events. Show that A c B if and only if A n B = A.
"If" part: We show that if A n B = A, then A c B. Let s E A. Then s E (A n B), since A = A n B.
Then by the definition of intersection, s E B. Therefore, A c B.
"Only if" part : We show that if A c B, then A n B = A. Note that from the definition of the intersec-
tion, (A n B) c A. Suppose s E A. If A c B, then s E B. So s E A and s E B; that is, s E (A n B). Therefore,
it follows that A c (A n B). Hence, A = A n B. This completes the proof.

1.15. Let A be an arbitrary event in S and let @ be the null event. Show that
(a) Au~=A
(b) AnD=0