JEE Advanced 2019 Paper 1 · Mathematics · Probability · hard

A particle starts at the point 3 on the set \0,1,2,…,8\. At each step it moves one unit to the right with probability 0.6 and one unit to the left…

Numerical answer

A particle starts at the point

3

on the set

\{0,1,2,\dots,8\}

. At each step it moves one unit to the right with probability

0.6

and one unit to the left with probability

0.4

, independently of the past. The walk stops as soon as the particle reaches

0

8

. Given that the walk stops at

8

, find the probability that its first step was to the left. Give the answer correct to two decimal places.

Show answer and solution

Answer: 0.32

This is a gambler's-ruin chain. With

\rho=q/p=\frac23

, the probability of ending at

8

from position

i

h_i=\frac{1-\rho^i}{1-\rho^8}

. Hence

h_3=\frac{1-8/27}{1-256/6561}\approx0.73228

and

h_2=\frac{1-4/9}{1-256/6561}\approx0.57811

. By the Markov property, after a left first step the particle is at

2

. Bayes' theorem gives

P(\text{left first}\mid\text{ends at }8)=\frac{0.4\,h_2}{h_3}=\frac{0.4\times0.57811}{0.73228}=\frac{6}{19}\approx0.3158

. The exact value is

\frac6{19}

, which rounds to

0.32

. Ignoring the conditioning would give

0.40

, and using

h_2

alone gives

0.58

Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake