Let A and B be events with peA) > 0 and PCB) > O. We say

QUESTION:

Let A and B be events with peA) > 0 and PCB) > O. We say that an event B suggests an event A if peA I B) > peA), and does not suggest event A if peA I B) < peA). (a) Show that B suggests A if and only if A suggests B. (b) Assume that P(BC) > O. Show t hat B suggests A if and only if BC does not suggest A. (c) We know that a treasure is located in one of two places, with probabilities f3 and 1 - f3, respectively, where 0 < f3 < 1. We search the first place and if the treasure is there, we find it with probability p > O. Show that the event of not finding the treasure in the first place suggests that the treasure is in the second place. Solution. (a) We have peA I B) = peA n B)/P(B), so B suggests A if and only if peA n B) > P(A)P(B), which is equivalent to A suggesting B, by symmetry. (b) Since PCB) + P(BC) = 1, we have P(B)P(A) + P(BC)P(A) = p eA) = P (B)P(A I B) + P(BC)P(A I BC), which implies that P(BC)(P(A) - peA I Be) = P(B) (P(A I B) - peA). P(A I s uggest A). A Using the > P(A) (8 B events A = treasure is B = { we t reasure in probability if P ( A) > P(B) = (AC)P(B I + (B I A) = t1( 1 - + (1 - ,6) ! so = ( A ) = ___ ____ = __-____ > 1 _ {3 = P ( B) .8( 1 - + ( 1 - /3) 1 - .8p . It follows that event B suggest event A.