Drunken man walking on an axis

Question

A drunken man walks randomly along an axis with the points $0, \pm 1, \pm 2, \ldots$. He begins from point zero and in each step he moves in a length of $1$ to the positive side with a probability of $0.4$ or to the negative side with a probability of $0.6$. The steps are independent. he does a total of 50 steps

What is the probability that he will be on the left of his starting point after $27$ steps?

I tried to calculate this using the binomial distribution with choosing $27$ out of $50$ using the given probabilities and I obtained $0.078$. Is this correct?

the way i'm trying to calculate is with$ \frac{\left(l+r\right)!}{l!r!}lp^lrp^r$ where l is the amount of left steps, r is the amount of right steps, lp is the probability of a left step, rp is the probability of the right step

Answer 1

Let $N$ be the random variable representing the number of steps he takes in the negative direction. We say $N\sim\mathrm B(27,0.4)$.

For him to end up on the negative side, he needs to take more steps to the negative than to the positive. In other words, he needs to take at least $14$ steps to the negative. Therefore, you need to calculate $\operatorname{P}(N\ge14)$.

Given that $$X\sim\mathrm B(n,p)\implies \operatorname{P} (X=x) = \binom nx p^x(1-p)^{n-x}$$ and that $$\operatorname{P}(a\le X\le b)=\sum_{x=a}^b \operatorname{P} (X=x)$$ can you continue?

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If you need to see the answer and think backwards, here it is:

$$\sum_{x=14}^{27} \binom{27}x(0.4)^x(0.6)^{27-x}$$

Answer 2

All you have to show is, after 27 steps, steps to the left is greater than steps to the right.

The probability of that is $27C14*0.6^{14}*0.4^{13} + 27C15*0.6^{15}*0.4^{12}.........27C27*0.6^{27}$. On a TI-83 or similar its 1 - binomcdf with an input of 27,0.6,13. That is, what is the probability of getting at least 14 lefts out of 27. I get close to p = .855