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I have a question about of random walk.

Consider a particle starting its random walk at 0. At each step, it either moves in positive or negative direction. If the probability of moving in positive direction is 0.6, what is the probability that it will end up at +1 after 5 steps?

I read this link. But I confused, because document says

  • List item If his first hop is to the left, then he lands at position n-1 and eventually falls into the Pit of Disaster with probability Rn-1.
  • On the other hand, if his first hop is to the right, then he lands at position n+1 and eventually falls into the Pit of Disaster with probability Rn+1.

Therefore, by the Total Probability Theorem, we have:

Total Probability Theorem

I couldn't understand how can I solve my question with these informations. Because this will give total probability. If you solve this question, I can understand. I don't need very detailed answer. Thanks

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Your question is simpler. Your net movement is $1$ to the right, one such situation is $LLRRR$, another is $LRRLR$,... so, you need $2L$ and $3R$ irrespective of the order. The probability of a particular $2L,3R$ move is $(0.6)^3(0.4)^2$. And, there are ${5 \choose 2}$ such situations, yielding ${5 \choose 2}(0.6)^3(0.4)^2$. It's the same as having $3$ heads in 5 coin tosses.

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