# HW3Answers(1.5-1.7) - exclusiveness P p ∞ u i =1 A i P =...

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Math331, Spring 2008 Instructor: David Anderson Section 1.5, 1.6 and 1.7 Hw answers Homework: pg. 34 #’s 1 , 3 , 4 , 5 , 10. pg. 36 #’s 5 , 9 , 11. pg. 34 problems: 1. The person will wait at least 10 minutes if the bus arrives in the time interval 1:10 - 1:30. Using the main deFnition of section 1.7, we see that the probability of this happening is 20 / 30 = 2 / 3. 3. Both are false. A perfectly good example is all of section 1.6. 4. P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 2 - P ( A B ). Also, P ( A B ) 1 (actually it’s equal to one and this can be shown because either A or B is a subset of A B , but I’ll continue with my proof). Therefore, 1 P ( A B ) = 2 - P ( A B ) and so P ( A B ) 2 - 1 = 1. But all probabilities are bounded above by 1 and so 1 P ( A B ) 1 and so P ( A B ) = 1. 5. We know that the probability that it is any speciFc number is 0. Therefore, let A i be the event that we choose the integer i ∈ { 1 , . . ., 1999 } . Thus, P ( A i ) = 0 and by the mutual exclusiveness of the sets, P p 1999 u i =1 A i P = 1999 s i =1 P ( A i ) = 1999 s i =1 0 = 0 . 10. As in the previous problem, we know that the probability of selecting any speciFc point is zero. The rationals are countable. Therefore, let { A i } i =1 represent the rationals in the interval (0 , 1). The probability that a number chosen is rational is therefore (by mutual

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Unformatted text preview: exclusiveness) P p ∞ u i =1 A i P = ∞ s i =1 P ( A i ) = 0 . If A is the event of choosing a rational and B is the event of choosing an irrational number, then A c = B and so P ( B ) = P ( A c ) = 1-P ( A ) = 1. pg. 36 problems: 5. The sample space consists of sequences of the form HTHTHTHT. .. or THTHTHTH until it ends with HH or TT , at which point it terminates. Thus we have HH , TT , HTT , THH , HTHH , THTT , . . . , ∈ S . To make this explicit, I would say: S = { a 1 a 2 ··· a k : k ∈ { 2 , 3 , . . . } , a i ∈ { H, T } , a i-1 n = a i for any i < k, a k = a k-1 } . 9. Let A be the red wine group and B be the white wine group. Then, P ( B ) = . 4 and P ( A ) = . 5 and P ( A ∪ B ) = . 7. We want P ( A ∩ B ) = P ( A )+ P ( B )-P ( A ∪ B ) = . 5+ . 4-. 7 = . 2. 1 11. We want P ( A c B c ) in terms of P ( A ), P ( B ), and P ( AB ). We have by De Morgan P ( A c B c ) = P (( A ∪ B ) c ) = 1-P ( A ∪ B ) = 1-P ( A )-P ( B ) + P ( AB ) . 2...
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HW3Answers(1.5-1.7) - exclusiveness P p ∞ u i =1 A i P =...

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