Page 24 - Schaum's Outlines - Probability, Random Variables And Random Processes

CHAP. 11 PROBABILITY

Shaded region: A n B Shaded region: A n B
Fig. 1-8

1.22. Let P(A) = 0.9 and P(B) = 0.8. Show that P(A n B) 2 0.7.
From Eq. (l.29), we have

P(A n B) = P(A) + P(B) - P(A u B)
By Eq. (l.32), 0 I P(A u B) I Hence
1.
P(A r\ B) 2 P(A) + P(B) - 1
Substituting the given values of P(A) and P(B) in Eq. (1.63), we get
P(A n B) 2 0.9 + 0.8 - 1 = 0.7
Equation (1.63) is known as Bonferroni's inequality.

1.23. Show that
P(A) = P(A n B) + P(A n B)
From the Venn diagram of Fig. 1-9, we see that
A = (A n B) u (A n B) and (A n B) n (A n B) = 0
Thus, by axiom 3, we have
P(A) = P(A n B) + P(A n B)

AnB AnB
Fig. 1-9

1.24. Given that P(A) = 0.9, P(B) = 0.8, and P(A n B) = 0.75, find (a) P(A u B); (b) P(A n B); and (c)
P(A n B).

(a) By Eq. (1 .29), we have
P(A u B) = P(A) + P(B) - P(A n B) =: 0.9 + 0.8 - 0.75 = 0.95
(b) By Eq. (1.64) (Prob. 1.23), we have
P(A n B) = P(A) - P(A n B) = 0.9 - 0.75 = 0.15
(c) By De Morgan's law, Eq. (1.14), and Eq. (1.25) and using the result from part (a), we get
P(A n B) = P(A u B) = 1 - P(A u B) = 1 - 0.95 = 0.05