probability of $1$ step fordward and $1$ step backward of drunked man.

A drunkard man takes a step forward with Probability of $0.6$ and takes a step backward with

Probability of $0.4.$ If he takes $9$ steps in all .Then the Probability that he is just one

step away from initial point.

$\bf{My\; Try}::$ here I have formed two cases

$\bullet$ If man take first step in fordward direction, Then he must take $5$ steps fordward and $4$ steps backwards so that he is one step away from initial point.

$\bullet$ If man take first step in backward direction, Then he must take $5$ steps backward and $4$ steps fordward so that he is one step away from initial point.

So probability for first cases:: $\displaystyle \bf{\binom{9}{5}}\cdot (0.6)^5\cdot (0.4)^4$

Similarly probability for second cases:: $\displaystyle \bf{\binom{9}{4}}\cdot (0.6)^4\cdot (0.4)^5$

So Total probability $\displaystyle \bf{\binom{9}{5}}\cdot (0.6)^5\cdot (0.4)^4+\bf{\binom{9}{4}}\cdot (0.6)^4\cdot (0.4)^5 = \binom{9}{5}\cdot (0.6)^4\cdot (0.4)^4\cdot 1$

Is my solution is Right. If not Then How can we solve it.

Help Required

Thanks

• Your answer is correct, but the reasoning isn't perfect: the two cases are whether his final position is one step forward or backward from his initial point (and have the probabilities you calculated). – ShreevatsaR Mar 19 '14 at 15:52