# The rst line of the answer is easy to see since p x 0

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Unformatted text preview: Using the multiplication rule this deﬁnition can be written in a more symmetric way as P (A \ B ) = P (A) · P (B ) (A.6) Example A.6. Roll two dice and let A = “the ﬁrst die is 4.” Let B1 = “the second die is 2.” This satisﬁes our intuitive notion of independence since the outcome of the ﬁrst dice roll has nothing to do with that of the second. To check independence from (A.6), we note that P (B1 ) = 1/6 while the intersection A \ B1 = {(4, 2)} has probability 1/36. P (A \ B1 ) = 1 14 6= · = P (A)P (B1 ) 36 6 36 210 APPENDIX A. REVIEW OF PROBABILITY Let B2 = “the sum of the two dice is 3.” The events A and B2 are disjoint, so they cannot be independent: P (A \ B2 ) = 0 &lt; P (A)P (B2 ) Let B3 = “the sum of the two dice is 9.” This time the occurrence of A enhances the probability of B3 , i.e., P (B3 |A) = 1/6 &gt; 4/36 = P (B3 ), so the two events are not independent. To check that this claim using (A.6), we note that (A.4) implies P (A \ B3 ) = P (A)P (B3 |A) &gt; P (A)P (B3 ) Let B4 = “the sum of the two dice is 7.” S...
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