Question

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 of travel. The second leads to a tunnel that returns him to his cell after three days of travel. The third door leads immediately to freedom. (a) Assuming that the prisoner will always select doors 1, 2 and 3 with probabilities 0.5, 0.3, 0.2 (respectively), what is the expected number of days until he reaches freedom

Answer 1

Answer:

10.5 days

Step-by-step explanation:

Given :

P(d1) = 0.5 ; P(d2) = 0.3 ; P(d3) = 0.2

For d1: 2days of freedom ; then back to cell

For d2 : 3 days of freedom ; then back to cell

For d3 :, complete freedom.

Let number of days till freedom is attained = x

x = E(x | d1)p(d1) + E(x | d2)p(d2) + E(x | d3)p(d3)

x = (2 + x)(0.5) + (3 + x)(0.3) + (1 * 0.2)

x = 1 + 0.5x + 0.9 + 0.3x + 0.2

x = 2.1 + 0.8x

x - 0.8x = 2.1

0.2x = 2.1

x = 2.1 / 0.2

x = 10.5 days

Hence, expected number of days to attain freedom = 10.5 days