a) A prisoner is trapped in a cell containing three doors. The first door leads to a tunnel that returns him to his cell after two days’ travel. The second leads to a tunnel that returns him to his cell after four days’ travel. The third door leads to freedom after one day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .3, .5, and .2, what is the expected number of days until the prisoner reaches freedom?
Answers
refer to the attachment below
Given : A prisoner is trapped in a cell containing three doors
The first door leads to a tunnel that returns him to his cell after two days’ travel. The second leads to a tunnel that returns him to his cell after four days’ travel. The third door leads to freedom after one day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .3, .5, and .2,
To Find : what is the expected number of days until the prisoner reaches freedom?
Solution:
Number of ways of selecting doors = 3! = 6
The third door leads to freedom after one day of travel.
Possible Actually as if he choose 3rd , he ii out
1 2 3 1 2 3
1 3 2 1 3
2 1 3 2 1 3
2 3 1 2 3
3 1 2 3
3 2 1 3
1 2 3 = (0.3)2 + (0.5)4 + (0.2)1 = 0.6 + 2 + 0.2 = 2.8
1 3 = (0.3)2 + (0.2)1 = 0.6 + 0.2 = 0.8
2 1 3 = (0.5)4 + (0.3)2 + (0.2)1 = 2 + 0.6 + 0.2 = 2.8
2 3 = (0.5)4 + (0.2)1 = 2 + 0.2 = 2.2
3 = (0.2)1 = 0.2
3 = (0.2)1 = 0.2
2.8 + 0.8 + 2.8 + 2.2 + 0.2 + 0.2 = 9 Days
expected number of days = 9
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