Question: A Building contains twoelevators, one fast and one slow. The average w

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Question: A Building contains twoelevators, one fast and one slow. The average waiting timefor the slow elevator is 3 minutes and the average waiting time ofthe fast elevator is 1 minute. If a passenger chooses thefast elevator with probability 2/3, and the slow elevator withprobability 1/3, what is the expected waiting time? (Use thelaw of total expectation, Markov's inequality theorem, definingappropriate random variables X and Y.) What I know:
We need conditional expectations for x=0 and x=1 and theunconditional pmf of X (elevator chosen). I do not think weknow the pdf of Y.
I am way confused. Please help me answer thisquestion

Explanation / Answer

Waiting time = Probability chooses fast lift *waiting time fast lift + probability choose slow lift* waiting timeslow lift now expected waiting time can be directlt calculated by takingthe expectation = E(Waiting time) = E( Probability chooses fast lift* waiting time fast lift + probability choose slow lift* waitingtime slow lift) = Probability chooses fast lift * E(waiting timefast lift) + probability choose slow lift* E(waiting time slowlift) as all four of the above quantities are known = 2/3*1 + 1/3 *3 = 5/3 minutes In terms of random variables Z is the final required randomvariable as waiting time = P(z)= 2/3 for Z =1 P(z) = 1/3 for Z = 3

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