HW2_Solution_2019.pdf - ENGG2470A Probability for Engineers Spring 2019 Solution for homework Two Solution to Problem 18(10\u2019 First we define the

HW2_Solution_2019.pdf - ENGG2470A Probability for Engineers...

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Spring 2019Solution for homework TwoSolution to Problem 18 (10’)First we define the following events,1s={The bit is 1 when storing},0s={The bit is 0 when storing},1d={The bit is decoded as 1},0d={The bit is decoded as 0}.Then according to Bayes’ Rule, we haveP(1s|1d)=P(1s)P(1d|1s)P(1d)=P(1s)P(1d|1s)P(1d|0s)P(0s) +P(1d|1s)P(1s)=0.5×0.85(1-0.9)×0.5 + 0.85×0.5= 0.8947.(1)Solution to Problem 23 (10’)First we define the following events,Qs={The candidate is qualified},Us={The candidate is unqualified},Qd={The candidate is identified to be qualified},Ud={The candidate is identified to be unqualified},A={20 questions will correctly identify someone to be qualified or unqualified}.Note thatP(Qs) =q, P(Us) = 1-q. In addition, denotex=P(Qd|Qs) =20Xk=1520kpk(1-p)20-k,(2)y=P(Qd|Us) =20Xk=1520k(1-p)kp20-k.(3)Then conditioning on whether the candidate is qualified or unqualified, we can apply the Total ProbabilityENGG2470A: Probability for Engineers
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Theorem to getP(A)=P(A|Qs)P(Qs) +P(A|Us)P(Us)=P(Qd|Qs)P(Qs) +P(Ud|Us)P
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  • Spring '18
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