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At each stage a person either moves one step to the right with probability 0.6 or one step to the left with probability 0.4.Assuming the direction of each step is independent

I want to find the expected number of steps it takes a person to be r steps to the right of where she began.

Answer:-15r steps. Is this right answer? If wrong, what would be correct answer?

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  • $\begingroup$ Use a random variable with binomial distribution where the outcome $1$ stands for "step to the right". Then you can use the fact that the sum of independent binomially distributed random variables again follows a binomial distribution. $\endgroup$
    – Dr_Be
    Jan 14, 2016 at 16:39
  • $\begingroup$ Sorry, was to slow in editing my previous comment which is not true! Next try from my side: Consider the r.v. $X_n$ which gives the position after $n$ steps. $n$ can be written as $n=k+(n-k)$ with $k$ equal to the number of steps to the right. The position is thus $X_n=k-(n-k)=2k-n$. What can be said about the distribution of $X_n$? $\endgroup$
    – Dr_Be
    Jan 14, 2016 at 16:46

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The answer should be 5 r.

The simple explanation is that for every 10 steps, the person takes (on average) 6 steps right and 4 steps left. Thus he moves (on average) 2 steps right / 10 steps or 1 step right per 5 steps.

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  • $\begingroup$ I was also thinking about the same answer.However,thanks. $\endgroup$ Jan 16, 2016 at 14:13

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