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hw2sol - 36-225 Homework 2 SOLUTIONS 1 15 Points 3 for each...

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36-225 - Homework 2 SOLUTIONS 1. 15 Points: 3 for each part (2 for work, 1 for numerical accuracy) If two events, A and B, are such that P(A) = .5, P(B) = .3, and P(A B) = .1, find the following: (a) P ( A | B ) = P ( A B ) P ( B ) = . 1 . 3 = 1 3 (b) P ( B | A ) = P ( A B ) P ( A ) = . 1 . 5 = 1 5 (c) P ( A | A B ) = P [ A ( A B )] P ( A B ) = P ( A ) P ( A ) + P ( B ) P ( A B ) = . 5 . 5 + . 3 . 1 = 5 7 (d) P ( A | A B ) = P [ A ( A B )] P ( A B ) = P ( A B ) P ( A B ) = 1 (e) P ( A B | A B ) = P [( A B ) ( A B )] P ( A B ) = P ( A B ) P ( A B ) = . 1 . 5 + . 3 . 1 = 1 7 2. 12 points: 2 for each part A smoke detector system uses two devices, A and B. If smoke is present, the probability that it will be detected by device A is .95; by device B, .90; and by both devices, .88. Here is the Venn Diagram for this problem: A ( . 95) B (.90) .07 .02 .88 .03 (a) If smoke is present, find the probability that the smoke will be detected by either device A or B or both devices. P ( A B ) = P ( A ) + P ( B ) P ( A B ) = . 95 + . 90 . 88 = . 97 (b) Find the probability that the smoke will be undetected. P(smoke undetected) = 1 P ( A B ) = 1 . 97 = . 03 (c) What is the probability that if smoke is present, it will be detected by exactly one of the two devices? P [( A B ) P ( A B )] = P ( A B )+ P ( A B ) = [ P ( A ) P ( A B )]+[ P ( B ) P ( A B )] = . 07 + . 02 = . 09 1
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(d) Given that smoke has been detected by device B, what is the probability that it will also be detected by device A? P ( A | B ) = P ( A B ) P ( B ) = . 88 . 90 = . 978 (e) If the smoke has not been detected by device B, what is the probability that it will be detected by device A? P ( A | B ) = P ( B A ) P ( B ) = . 07 . 10 = . 7 (f) No, the two devices are not independent! In order to show this, we need to pick one of the four conditions for independence presented in class and show that it does not hold. In particular, you can use on of the following: P ( A B ) negationslash = P ( A ) P ( B ) [ . 88 negationslash = . 95 × . 90] P ( A | B ) negationslash = P ( A ) [from part (d), . 978 negationslash = . 95] P ( A | B ) negationslash = P ( A | B ) [from parts (d) and (e) . 978 negationslash = . 70] 3. 5 points P ( ¯ A ¯ B ) = a P ( B ) = b Need to find P ( A ). P ( ¯ A ¯ B ) = a = P ( A B ) = 1 a Using the addition rule, and the fact that A and B are independent, we can setup the following equation: 1 a = P ( A ) + P ( B ) P ( A ) P ( B ) = P ( A ) + b P ( A ) · b = (1 b ) · P ( A ) + b This implies that: P ( A ) = 1 a b 1 b 4.
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