A miner is trapped in mine containing 3 doors The first door will take him out of the tunnel in 2 hours the second door will take hin out of the tunnel in 5 hours and the third door will take him out of the tunnel in 7 hours. Assuming that the miner at all times is equally likely to choose any of the doors, what is the expected length of time until he reaches safety.(adsbygoogle = window.adsbygoogle || []).push({});

Let X denote the time he reaches safety, Y denotes the initial choice

E(X) = 1/3 [E(X|Y=1)+E(X|Y=2)+E(X|Y=3)]

E(X|Y=1) = 2

E(X|Y=2) = 5 + E(X) <- I DONNO WHY IT ADDS E(X)

E(X|Y=3) = 7 + E(X) <- AGAIN...DONNO WHY...

E(X) = 1/3[14+2E(X)]

THUS, E(X) = 14.

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# Homework Help: A probability question!

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