document.write( "Question 441203: 3.18 Suppose A and B are events with 0 < P(A) < 1 and 0 < P(B) < 1.
\n" ); document.write( "a. If A and B are disjoint, can they be independent?
\n" ); document.write( "b. If A and B are independent, can they be disjoint?
\n" ); document.write( "c. If A ⊂ B, can A and B be independent?
\n" ); document.write( "d. If A and B are independent, can A and A ∪ B be independent
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Algebra.Com's Answer #304541 by robertb(5830)\"\" \"About 
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a. A, B disjoint means P(A and B) = 0. Since P(A), P(B) > 0, \"P%28A%29%2AP%28B%29+%3E+0\", hence \"P%28A%29%2AP%28B%29+%3C%3E+P%28A+and+B%29\", and so are not independent.\r
\n" ); document.write( "\n" ); document.write( "b. A, B independent means \"P%28A%29%2AP%28B%29+=+P%28A+and+B%29\"
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\n" ); document.write( "Unless P(A)*P(B)= 0, A and B can never be disjoint. But from the given, P(A)*P(B) > 0, hence A, B are never disjoint.\r
\n" ); document.write( "\n" ); document.write( "c. If A ⊂ B, then P(A and B) = P(A). If we assume independence, then
\n" ); document.write( "P(A) = P(A)*P(B), which means
\n" ); document.write( "P(B) = 1, a contradiction of the hypothesis that 0 < P(B) < 1. Hence A, B are not independent.\r
\n" ); document.write( "\n" ); document.write( "d. P(A and (A ∪ B)) = P(A), since A ⊂ A U B. Hence the argument is reduced to one like in part (c) above, the result being the same.
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