hw2sol - 36-225 - Homework 2 SOLUTIONS1.15 Points: 3 for...

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Unformatted text preview: 36-225 - Homework 2 SOLUTIONS1.15 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(AB) = .1, find thefollowing:(a)P(A|B) =P(AB)P(B)=.1.3=13(b)P(B|A) =P(AB)P(A)=.1.5=15(c)P(A|AB) =P[A(AB)]P(AB)=P(A)P(A) +P(B)P(AB)=.5.5 +.3.1=57(d)P(A|AB) =P[A(AB)]P(AB)=P(AB)P(AB)= 1(e)P(AB|AB) =P[(AB)(AB)]P(AB)=P(AB)P(AB)=.1.5 +.3.1=172.12 points: 2 for each partA smoke detector system uses two devices, A and B. If smoke is present, the probability thatit 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 deviceA or B or both devices.P(AB) =P(A) +P(B)P(AB) =.95 +.90.88 =.97(b) Find the probability that the smoke will be undetected.P(smoke undetected) = 1P(AB) = 1.97 =.03(c) What is the probability that if smoke is present, it will be detected by exactly one ofthe two devices?P[(AB)P(AB)] =P(AB)+P(AB) = [P(A)P(AB)]+[P(B)P(AB)] =.07 +.02 =.091(d) Given that smoke has been detected by device B, what is the probability that it will alsobe detected by device A?P(A|B) =P(AB)P(B)=.88.90=.978(e) If the smoke hasnotbeen detected by device B, what is the probability that it will bedetected by device A?P(A|B) =P(BA)P(B)=.07.10=.7(f) No, the two devices arenot independent!In order to show this, we need to pick one of the four conditions for independencepresented in class and show that it doesnothold. In particular, you can use on of thefollowing:P(AB)negationslash=P(A)P(B) [.88negationslash=.95.90]P(A|B)negationslash=P(A) [from part (d),.978negationslash=.95]P(A|B)negationslash=P(A|B) [from parts (d) and (e).978negationslash=.70]3.5 pointsP(AB) =aP(B) =bNeed to findP(A).P(AB) =a=P(AB) = 1aUsing the addition rule, and the fact thatAandBare independent, we can setup the followingequation:1a=P(A) +P(B)P(A)P(B) =P(A) +bP(A)b= (1b)P(A) +bThis implies that:P(A) =1ab1b4.12 points: 6 for each part: 2 for setup, 3 for work, 1 for numerical accuracyLetRi= relayiis openi= 1,2,3,4Design AP(current flows)=P{(R1R2)(R3R4)}independencebracehtipdownleftbracehtipuprightbracehtipupleftbracehtipdownright=P(R1R2) P(R3R4)={P(...
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hw2sol - 36-225 - Homework 2 SOLUTIONS1.15 Points: 3 for...

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