Question

trapped in a cell that contains 3 doors. The first door leads to a tunnel that returns her to back tothe cell after 2 days of travel. The second door leads to a tunnel that returns her back to the cell after 4days of travel. The third door leads her freedom after 1 day of travel. If it is assumed that Ally will alwaysselect doors 1, 2, and 3 with probabilities 0.5, 0.3, and 0.2 (respectively), what is the expected number ofdays until she reaches freedom

Answer 1

Answer:

The expected number of days until prisoner reaches freedom is 12 days

Step-by-step explanation:

From the given information:

Let X be the random variable that denotes the number of days until the prisoner reaches freedom.

We can evaluate E(X) by calculating the doors selected, If Y be the event that the prisoner selects a door, Then;

E(X) = E( E[X|Y] )

E(X) = E [X|Y =1 ] P{Y =1} + E [X|Y =2 ] P{Y =2} + E [X|Y =3 ] P{Y =3}

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

$E(X) = (2 + E[X])0.5+ (4 + E[X])0.3+ 0.2$

Solving for E[X]; we get

E[X] = 12