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# Science 2

## Question 1:

We know a random walk of $ n $ steps by any of the drunk guys is a random sequence of $ L's $ and $ R's $.

Since they have to meet after n steps, they make 2N steps together.
For e.g, $ LLRRLRRL... $ is such a sequence.

Total number of ways to make exactly $ l $ left steps out of total $ n $ steps (and $ n-l $ right steps) = $ \large{n \choose l} $

Now for both of the drunk guys to meet at same place after $ n $ steps, they both need to make exactly same number of steps towards left and right.

Total number of ways for both of the drunk guys to make exactly $ l $ steps towards left = $ \large{n \choose l}^{2} $

Total number of ways for both drunk guys to meet together = $\large{\sum_{l=0}^{n}{n\choose l}^2}$ = $ {2n \choose l} $

Total number of possible random walks for each guy = $ 2^n $

Total number of possible random walks for both guy together = $ 2^n * 2^n $ = $ 4^n $

Probabilty that both the drunk guys meet after walking $ N $ random steps = $\large{\frac{{2n \choose l}}{4^n}}$ $ (ans)$

### Related question 1

For, the random walk to end at centre, the walk should consist of equal number of left and right steps.

Thus if $ n $ is $ even $, the number of such possible walks = $ \large{n \choose n/2} $

With an $ odd $ $ n $ , the number of such walks = $ 0 $

Total number of random walks of $ n $ steps = $ 2^n $

Thus, if $ n $ is $ odd $ the probabilty of random walk to end at centre = $ 0 $

If $ n $ is $ even $ the probabilty of random walk to end at centre = $\large{\frac{{n \choose n/2}}{2^n}}$ $ (ans)$

### Related question 2

Let's define,
$ X_i $ be the random variable denoting the $ ith $ step out of $ n $ steps.

Thus, $ X_i $ = $ -1 $ if the $ ith $ step is toward $ left $ and $ X_i $ = $ 1 $ if $ ith $ step is toward $ right $.

Therefore final displacement after $ n $ steps, $ d $ = ${\sum_{i=0}^{n} {X_i}} $.

$ \implies{} Ed = {\sum_{i=0}^{n} E{X_i}} $

Now, $ EX_i = \frac{{1}}{2} * (1) + \frac{{1}}{2} * (-1) = 0 $

$ \implies{} Ed = n*0 = 0 $ $(ans)$

### Related question 3

Let's define,
$ X_i $ be the random variable denoting the $ ith $ step out of $ n $ steps.

Thus, $ X_i $ = $ -1 $ if the $ ith $ step is toward $ left $ and $ X_i $ = $ 1 $ if $ ith $ step is toward $ right $.

Therefore final displacement after $ n $ steps, $ d $ = ${\sum*{i=0}^{n} {X_i}} $.
$ \implies{} d^2 = (\sum*{i=0}^{n} {X

*i} )^2$*

$ \implies{} d^2 = (X_1^2 + X_2^2 + ... + X_n^2) + 2(X_1X_2 + X_1X_3 + ... + X{n-1}X

$ \implies{} d^2 = (X_1^2 + X_2^2 + ... + X_n^2) + 2(X_1X_2 + X_1X_3 + ... + X

*{n})$*

$ \implies{} d^2 = \sum{i=0}^{n} {X

$ \implies{} d^2 = \sum

*i}^2 + 2\sum*{i\neq{}j,i=0,j=0}^{n} {X_iX_j}$

$ \implies{} Ed^2 = \sum*{i=0}^{n} E{X_i}^2 + 2\sum*{i\neq{}j,i=0,j=0}^{n} E{X_iX_j}$

Now, $ EX_i^2 = \frac{{1}}{2} * (1)^2 + \frac{{1}}{2} * (-1)^2 = 1$

Similarly, $ EX_iX_j = \frac{{1}}{4} * (1* 1) + \frac{{1}}{4} * (-1* 1) + \frac{{1}}{4} * (1* -1) + \frac{{1}}{4} * (-1* -1) = 0$

$ \implies{} Ed^2 = n*1 + 0 = n$.

Thus the mean-squared displacement after $ n $ steps = $ n $ $ (ans) $