Question

Exercise 6: A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 hours of travel. The second door leads to a tunnel that will return him to the mine after 5 hours of travel. The third door leads to a tunnel that will return him to the mine after 7 hours. If we assume that the miner is at all times equally likely to choose any one of the doors, what is the expected length of time until he reaches safety? Can you also try to find the variance of length of time until he reaches safety?

Answer 1

The mine contains 3 doors.

The first door will take the miner to safety in 3 hours.

The second door will take the miner to safety in 5 hours.

The third door will take the miner to safety in 7 hours.

The choice of 3 doors is equally likely,

ie. the miner chooses door 1,door 2 and door 3 ,each with probability 1/3.

Let,X be the random variable denoting the time he requires to reach safety.

(a) To find  E(X)

E(X)

=(3+5+7)/3

=5

So,the expected time the miner takes to reach safety is 5 hours.

(b) To find Var(X)

Now,

E(X²)

=(3²+5²+7²)/3

=83/3

=27.66

So,

Var(X)

=27.66-5²

=2.667

So,the variance of the length of time until he reaches safety is

2.667.