Question

A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days’ travel. The second leads to a tunnel that returns him to his cell after 4 days’ travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .5, .3, and .2, what is the expected number of days until the prisoner reaches freedom?

Answer 1

Answer:

12 days

Step-by-step explanation:

$E[X]=E[E[X|Y]]$

$E[E|Y=1]P{Y=1}+E[E|Y=2]P{Y=2}+E[E|Y=3]P{Y=3}$

$(2+E[X])\frac{1}{2} +(4+E[X])\frac{3}{10}+1(\frac{2}{10} )$

Solving for E(x) we get 12