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 0.3, 0.5, and 0.2, what is the expected number of days until the prisoner reaches freedom?

Question

Mathematics

Rodney

22 April, 23:18

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 0.3, 0.5, and 0.2, what is the expected number of days until the prisoner reaches freedom?

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Jimmy Sawyer · Answer 1

The expected number of days until the prisoner reaches freedom is 2.8.

Step-by-step explanation:

Door 1: 0.3 probability of being selected. Leads to his cell after two days' travel.

Door 2: 0.5 probability of being selected. Leads to his cell after four days' travel.

Door 3: 0.2 probability of being selected. Leads to his cell after one day of travel.

What is the expected number of days until the prisoner reaches freedom?

We multiply the probability of each door being used by the time that it leads to the cell. So

E = 0.3*2 + 0.5*4 + 0.2*1 = 2.8

The expected number of days until the prisoner reaches freedom is 2.8.