IIT JEE , AIEEE Forum - Mathematics , Algebra

ramkumar_november

A Man may take a step forward,backward left or right with equal probability. Find the probability that after nine steps he will be just one step away from his initial position.

please give detailed solution........... rates assured...........

feynmann

{ 9c4 + 9c5 } /2^9

a simple binomial disn.

chimanshu_007

2 conditions....

man will take 5 steps forward and 4 backward.....

or he will take 4 steps forward and 5 backward

x = 1 step forward = 0.5

y = 1 step backward = 0.5

req probability = ⁹C₅. x⁵.y⁴ + ⁹C₄ . x⁴.y⁵

= ⁹C₄ (x⁵.y⁴ + x⁴.y⁵)

= ⁹C₄((0.5)⁹ + (0.5)⁹)

= 126 . 2 (0.5)⁹

solve this now.....

ramkumar_november

feynmann and himanshu 's answers are wrong.........
and himanshu ...... there are more than 2 conditions ...............
if the man moves one step forward from there he can move in four directions ... you have assumed only one direction .......

akhil_o

PS not sure.. tell me if theres any mistake

If we plot the points possible to reach from initial position, we get a tilted square.U can try by taking lesser than 9 first, for eg. 2 or 3.We get a diagram as shown below.The dots represent possible positions after 9 steps, the darkest dot is the starting position.Steps are taken along 1 diagonal

Total points are 2*(1+3+5+...17)+ 19-1

(We subtract 1 as we can't reach startin position in odd no of steps)

=180.

4 points are a distance of 1 step from I.P

therefore probability=4/180=1/45

PS. I am not sure of the solution...if u find any mistakes pl reply and do not hesitate to vote!

venky1907

well done boys ... :)

ramkumar_november

Good try akhil ......... but the answer given is 3969/4⁷

please someone try to prove it...............

konichiwa2x

I'll post the explanation tomorow... very nice question by the way...

rv_hbk

p(forward)=0.5,p(backward)=0.5

At the end of 9 steps, to be i step away from the starting pt.,

I (no. of forward)-(no.of backward) I=1;

So, probability= ⁹C₅*(0.5)⁴* (0.5)⁵+ ⁹C₄*(0.5)⁵* (0.5)⁴

⁼ (⁹C₅+ ⁹C₄)*(0.5)⁹

=252/512 (ANS)

nadeemoidu

A Man may take a step forward,backward , left or right with equal probability. Find the probability that after nine steps he will be just one step away from his initial position.

Pls don't post any more answers taking only the cases of forward and backward!!!!

@akhil It is not necessary that he should move 4 steps and then come back . he can just move 1 step , then come back 1 step , again go 1 step , come back 1 step , etc.

konichiwa2x

This is a random walk on the lattice of integer points in the plane. At each stage, there is a probability 1/4 of moving to each of the four adjacent points. There are

possible paths (starting at the origin) of length n. The number of these that end at the point (j,k) is 0 if

or if

is odd. If

is even then the number of paths ending at (j,k) is

.

I guessed that formula by working out what happens for small values of n (2,3,4 and 5). My iit proffessor proved it by induction on n, by using the recurrence relation ,

where N(n;j,k) is the number of paths of length n ending at (j,k). It follows from the standard property of binomial coefficients,

.

So, the probability that the man is one step away from where he started after 9 steps is

.

ps. i am extremely sorry ramkumar_november, for the late reply. my board practicals were going on so didnt have time.

ramkumar_november

thanks a lot konichiwa......... i was not able to decipher that formula since i dont know much about recurrence......... any way do you think iit may ask such type of questions???????

konichiwa2x

no i dont think so...this is definitely beyond jee level...i was able to complete it only with my sir's help and he says the same thing.

Ask & Discuss Questions with Community & Experts