Question

A boy stands on a street corner and tosses a fair die. If 1 or 2 occurs, he walks one block west; otherwise he walks one block east. At his new position he repeats the procedure. What is the probability that after having tossed the coin 6 times he is

(a)At his starting point

(b)2 blocks from his starting point;

(c)4 blocks from his starting point.

Answer 1

The die has 6 possible outcomes. Then the probability of going west is 2/6 = 1/3, and the probability of going east is 4/6 = 2/3.

(a) To end up at the starting point, the boy needs to have an equal number of east and west walks which is 3 of each. Different die tosses are independent, so probabilities are multipled. Then the probability is

$\displaystyle P(A) = \Big(\frac{1}{3} \Big)^3 \cdot \Big (\frac{2}{3}\Big)^3 = \frac{2^3}{3^6}=\frac{8}{729} =0.011 = 1.1\%$

(b) In this case, the boy must walk 2 times east, 2 times west (to come to starting point) and either 2 east or 2 west. The order here doesn't matter, because we count the total displaysment which will be the same in any combination (w-e-e-w or e-w-w-e will give the same results). Because of the last condition, 2 east or 2 west, we need to add those probabilities.

$\displaystyle P(B) = \Big(\frac{1}{3} \Big)^2\cdot \Big (\frac{2}{3}\Big)^2 \cdot \Big(\frac{1}{3} \Big)^2 + \Big(\frac{1}{3} \Big)^2\cdot \Big (\frac{2}{3}\Big)^2 \cdot \Big(\frac{2}{3} \Big)^2 = \frac{4}{3^6}+ \frac{16}{3^6}=\frac{20}{729} =0.027 = 2.7\%$

(c) This time boys must walk 1 time east, 1 time west, and either 4 east or 4 west.

$\displaystyle P(C) = \frac{1}{3} \cdot \frac{2}{3} \cdot \Big(\frac{1}{3} \Big)^4 + \frac{1}{3} \cdot \frac{2}{3} \cdot \Big(\frac{2}{3} \Big)^4 = \frac{2}{3^6}+ \frac{2^5}{3^6}=\frac{34}{729} =0.047 = 4.7\%$

Answer: a) 1.1%, b)2.7%, c) 4.7%