03Summer-HE1-soln - University of Illinois 1(a First Hour...

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University First Hour Exam: Solutions ECE 313 of Illinois Page 1 of 2 Summer 2003 1(a). FALSE Conditional probabilities can be smaller than or equal to the unconditional probabilities. FALSE TRUE Conditional probabilities satisfy axioms of probability. TRUE The first term, P(B | A) P(A), is equal to P(A | B) P(B). This is just the theorem of total probability. FALSE This does not hold unless P(A) = P(B) or P(AB) = 0. (b) All four statements are true. A B A P(A B) P(A), and A B B P(A B) P(B). Thus, P(A B) min{P(A),P(B)}. Similarly, A A B P(A) P(A B), and B A B P(B) P(A B). Thus, P(A)+P(B) 2P(A B). Also, 1 P(A B) = P(A) + P(B) – P(A B) P(A B) P(A) + P(B) – 1. Finally, P(A | B) = P(A B)/P(B), since 0 < P(B) < 1, P(A | B) P(A B). 2. This problem is best solved with a Venn diagram or more preferably a Karnaugh map as shown below. From the given values of P(ABC) = 0.01, P(AB) = 0.08, P(BC) = 0.04, and P(AC) = 0.02, it is easy to fi nd P(ABC c ) = 0.08 – 0.01 = 0.07, P(A c BC) = 0.04 – 0.01 = 0.03, and P(AB c C) = 0.02 – 0.01 = 0.01. Next, since P(A) = 0.1 = P(ABC) + P(ABC c ) + P(AB c C) + P(AB c C c ), we get P(AB c C c ) = 0.01. Similarly, we can show that P(A c BC c ) = 0.19, and P(A c B c C) = 0 . Finally, we can get P(A B C) = P(ABC) + P(ABC c ) + P(A c BC) + P(AB
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