Question

A rat is trapped in a maze. Initially he has to choose one of two directions. If he goes to the right, then he will wander around in the maze for three minutes and will then return to his original position. If he goes to the left, then with probability 1/3 he will depart the maze after two minutes of travelling, and with probability 2/3 he will return to his initial position after 5 minutes of travelling. Assuming that the rat is at all times equally likely to go to the left or to the right, what is the expected number of minutes the rat will be trapped in the maze?

Answer 1

Answer:

The expected number of minutes the rat will be trapped in the maze is 21 minutes.

Step-by-step explanation:

The rat has two directions to leave the maze.

The probability of selecting any of the two directions is, $\frac{1}{2}$.

If the rat selects the right direction, the rat will return to the starting point after 3 minutes.

If the rat selects the left direction then the rat will leave the maze with probability $\frac{1}{3}$ after 2 minutes. And with probability $\frac{2}{3}$ the rat will return to the starting point after 5 minutes of wandering.

Let X = number of minutes the rat will be trapped in the maze.

Compute the expected value of X as follows:

$E(X)=[(3+E(X)\times\frac{1}{2} ]+[2\times\frac{1}{6} ]+[(5+E(X)\times\frac{2}{6} ]\\E(X)=\frac{3}{2} +\frac{E(X)}{2}+\frac{1}{3}+\frac{5}{3} +\frac{E(X)}{3} \\E(X)-\frac{E(X)}{2}-\frac{E(X)}{3}=\frac{3}{2} +\frac{1}{3}+\frac{5}{3} \\\frac{6E(X)-3E(X)-2E(X)}{6}=\frac{9+2+10}{6}\\\frac{E(X)}{6}=\frac{21}{6}\\E(X)=21$

Thus, the expected number of minutes the rat will be trapped in the maze is 21 minutes.