Probability that we pass through the park 
Assume that you live at the origin and your office is at $(4,5)$. What is the probability that you will pass through a park at $(2,4)$
  ? Given that you either move one block in $+x$ or $+y$ in one step and all paths are equally likely.

My Attempt:


Is my attempt correct? And also will their be an alternate solution to this question as my method may become quite cumbersome if we have a bigger grid?

 A: All possibilites to go through the park is $${6\choose 2}\times {3\choose 2} = 15\times 3 = 45$$
All pssible ways to come at office is $${9\choose 4} = 126$$
So the answer is $$P = {45\over 126 } = {5\over 14}$$
A: The total number of paths possible is $\binom{9}{5}$. (See why? Of the 9 steps you make, choosing when you will make the 5 y-steps determines the path entirely. Alternatively, you can choose when you will make the 4 x-steps). The number of paths through the park is $\binom{6}{2} \binom{3}{1}$. So the probability you need is $$\frac{ \binom{6}{2} \binom{3}{1}}{\binom{9}{5} }$$ 
A: To begin, show, by induction, that, on an m by n grid, there are $\frac{(m+n)!}{m!n!}$ paths from one corner to the diagonally opposite corner.  In the entire grid m= 4 and n= 5 so the number of paths from home to office is $\frac{9!}{4!5!}= 126$.  Now look at the 4 by 2 grid from home to park.  There are $\frac{6!}{2!4!}= 15$ paths that go to the park.
A: Your attempt is not correct, because the condition "All paths are equally likely" is not equivalent to the condition "You are equally likely to take one step north or one step each, when you have the option" (which is the condition you assumed). To see this, consider what your method gives as the probability of taking the path up the left side of the grid and then across the top (5 steps north then 4 steps east). You get $(\frac{1}{2})^5$ (because you have choices for the first 5 steps, but the last 4 steps are determined). Now consider what your method givesa s the probability of taking the path across the bottom and up the right (4 steps east then 5 steps north) -- you get $(\frac{1}{2})^4$ by the same method. So the probability of these two paths are not equal; thus according to your method, not all paths are equally likely.
The correct way to implement "All paths are equally likely" is by the following nice combinatorial trick: you know you will take $9$ steps, and you know $5$ of them will be north and $4$ east, so choosing is equivalent to arranging the letters NNNNNEEEE (since each arrangement will dictate a different path). This is equivalent to choosing $5$ spots among $9$ for Ns, so there are $\binom{9}{5}$ ways to do this. In general, by this method, there are $\binom{n+m}{n}$ paths which take $n$ steps north and $m$ steps east.
To caclulate paths that pass through the park, you need to multiply the number of paths from home to the park (4 steps north, 2 east) by the number of paths from the park to the office (1 step north, 2 east). So you get $\binom{6}{4}\binom{3}{1}$.
So the probability should be $\dfrac{\binom{6}{4}\binom{3}{1}}{\binom{9}{5}}$
