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Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 5: Solutions Conditional Probability, Law of Total Probability, Bayes Formula 1. [Picking a random subset] (a) There are n k different subsets of size k , and since each is equally likely to be drawn, the probability of drawing a specific subset is just n k 1 = k ( k 1) 2 1 n ( n 1) ( n k + 1) . (b) There are k desirable objects among the set of n objects and if we draw one at random, the probability that it is a desirable object is k/n , that is, P ( A 1 ) = k/n . If the event A 1 has occurred, then there are only k 1 desirable objects left among the n 1. Thus, P ( A 2  A 1 ) = ( k 1) / ( n 1). If both A 1 and A 2 have occurred, there are k 2 desirable objects left among the n 2 and so P ( A 3  A 1 A 2 ) = ( k 2) / ( n 2) and so on. More generally, P ( A i  A 1 A 2 A i 1 ) = ( k ( i 1)) / ( n ( i 1)) since i 1 desirable objects have been drawn already. Finally, P ( A 1 A 2 A k ) = P ( A 1 ) P ( A 2  A 1 ) P ( A k  A 1 A 2 A k 1 ) = k ( k 1) 2 1 n ( n 1) ( n k + 1) . 2. [The beginning of spring training] Since the pitcher can only pitch fast balls, curve balls, and sliders, we have that P ( F ) + P ( C ) + P ( S ) = 1, and since P ( C ) = 2 P ( F ), we conclude that P ( S ) = 1 3 P ( F ). The law of total probability gives P ( H ) = 1 4 = P ( H  F ) P ( F ) + P ( H  C ) P ( C ) + P ( H  S ) P ( S ) = P ( H  F ) P ( F ) + P ( H  C ) 2 P ( F ) + P ( H  S )(1 3 P ( F )) = 2 5 P ( F ) + 1 4 2 P ( F ) + 1 6 (1 3 P ( F )) from which we get that P ( F ) = 5 24 ,P ( C ) = 10 24 , and P ( S ) = 9 24 . 3. [Exercises in Conditional Probability] (a) P ( A  B c ) = 1 P ( A c  B c ) = 0 . 6. P ( A ) = P ( A  B ) P ( B ) + P ( A  B c ) P ( B c ) = 0 . 3 . 7 + 0 . 6 . 3 = 0 . 39. P ( B  A ) = P ( A  B ) P ( B ) /P ( A ) = 0 . 3 . 7 / . 39 = 7 / 13....
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 Spring '08
 Milenkovic,O

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