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A particle starting at the origin once every second has a 1/6 probability of jumping two steps forward, a 1/6 probability of jumping one steps forward, a 1/3 probability of jumping one step back, and a 1/3 probability of staying in the same place.

Calculate the probability that at the 36 second mark the particle is more than ten units away from the origen.

I've messed around with this problem and think the expected value is +6 units from the origen and the variance is 41 but I don't know what to do to actually figure out the probabilities here. Thanks.

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    $\begingroup$ Do you know whether this is expected to be done by hand or with a computer? Because by hand the only approach I can see that isn't hopelessly tedious is a normal approximation. For that normal approximation, all you need is the mean and variance that you just calculated. $\endgroup$
    – Ian
    Nov 22, 2019 at 20:50
  • $\begingroup$ Thanks. Definitely by hand. I actually had though of that + that was my response. My TA cautioned me that that wasn't the right approach but it was his first time looking at the problem and I think he was mistaken thinking that multinomial problem solving wouldn't be tedious. $\endgroup$
    – hotzjacobb
    Nov 22, 2019 at 21:00

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Unless someone has a better non-super intensive method; we use the Central limit theorem which states the successions of i.v. when the reach a large enough number tend to resemble a normal distribution. Thus the z-point value are .625 and -2.5 respectively and that results in a .006 prob of being in the left tail and a .266 prob. of being in the right tail which gives us .272 prob. of being 10 or more units away from the origin.

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