Question

A drunk person is walking on the road. With probability 0.6 he takes a step forward and with probability 0.4 he takes a step backward. After 10 steps, what is the probability that he is at his starting position?

Answer 1

If a person after 10 steps is at his starting position, then this person takes 5 steps forward and 5 steps backward.

This is a binomial distribution with

• probability to take step forward $p=0.6;$
• probability to take step backward $q=0.4;$
• $n=10.$

Then

$Pr(\text{5 steps forward and 5 steps backward})=C_{10}^5p^5q^5=\\ \\=\dfrac{10!}{5!(10-5)!}\cdot (0.6)^5\cdot (0.4)^5=\dfrac{10!}{5!\cdot 5!}\cdot (0.6\cdot 0.4)^5=\\ \\=\dfrac{5!\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10}{5!\cdot 2\cdot 3\cdot 4\cdot 5}\cdot (0.24)^5=7\cdot2\cdot 9\cdot 2\cdot (0.24)^5=252\cdot (0.24)^5=0.2006581248.$

Answer: $252\cdot (0.24)^5=0.2006581248.$