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**1. Let p<q where p is success (+1) and q is failure (-1). Let T=last time we hit 0 (T[tex]\geq[/tex]0) Find P(T=t)**

Then using the answer from the above question, make up a formula for [tex]\sum(2n choose n)*x^{n}[/tex]

I know the law of large numbers says that the random walk will eventually enter a downward cone and never leave, meaning that at a long enough time the amount of loses will be greater than the wins because of p and q. But I don't know where this gets me.

Then using the answer from the above question, make up a formula for [tex]\sum(2n choose n)*x^{n}[/tex]

## The Attempt at a Solution

I know the law of large numbers says that the random walk will eventually enter a downward cone and never leave, meaning that at a long enough time the amount of loses will be greater than the wins because of p and q. But I don't know where this gets me.