HW05 - University of Illinois Fall 2009 ECE 313: Problem...

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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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HW05 - University of Illinois Fall 2009 ECE 313: Problem...

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