Probability

#1
A hiker is lost in a cave, and he wants to get out of the cave. Three passages
are open to her. The first passage leads to the outside after 1 hour of travel.
The second passage winds around and will return her to the same place
after 3 hours of travel. The third and final passage also winds around and
will return her to the same place after 5 hours of travel. If the hiker is at all
time equally likely to choose any one of the passages, what is the expected
length of time until he reaches the outside?

I know its 33% for each route. But I have no idea how to determine the final hour amount.

One could spend 10 hours in there going down the same route twice. or, you could spend 1 hour in there if you chose properly the first time.

The final answer is 7 hrs. I just have no idea how to get there
 
#2
First, you must list all of the different ways she can reach outside together with their associated travel times and probabilities. (Hint: There are five of them when you assume she's not an idiot and won't choose the same passage a second time if it takes her back to the start.)

Once you have a list of all the possibilities, their overall probabilities and associated travel times, it's a trivial matter to calculate the expected value. I assume you understand the principle behind expectation.
 
#3
"If the hiker is at all time equally likely to choose any one of the passages"

If there were only 5 it wouldn't to bad, unfortunately, the hiker is allowed to be stupid. And could go down the same path 100 times before realizing that they went in circles
 
#4
Okay, in that case enumerate the first few travel times + probabilities for getting out after 1, 2, 3, … k attempts. (Hint: The first passage can only ever be the one chosen last.) Then calculate the composite expected time as k → ∞.

BTW, the final answer of 7 hours is not correct in either case where the hiker does or does not use the same passage.