Point Path Probability

Let $|P|$ denote the number of paths in $P$ . Additionally, note that there are $25 \times 25 = 625$ total points. Thus, since any pair $(s, p)$ is formed from an arbitrarily chosen $s$ and arbitrarily chosen $p$ , there are thus $625 \times |P|$ total possible pairs.

To count the number of pairs $(s, p)$ such that $s \in p$ , first note that any path must pass through the same number of points ( $25$ points up and $24$ across or $24$ points up and $25$ across depending on interpretation), which can be seen to be $25+24=49$ points. Thus, for any arbitrarily chosen $p$ , there are $49$ such $s$ such that $s \in p$ . So for any pair $(s, p)$ where $p$ is fixed, there are thus $49$ pairs satisfying the condition. Since this applies for any $p \in P$ , there are thus a total of $49 \times |P|$ pairs satisfying the condition.

The probability of picking a pair satisfying the condition is thus $\frac{49|P|}{625|P|} = \frac{49}{625}$ So $a=49$ , $b=625$ and therefore the answer is $a+b =49+625= \boxed{674}$ .

imagine that you have 25 x 25 grid. maybe too big.

imagine that you have 4 x 4 or 5 x 5 or anysize of grid you think easy, then give any color that match the condition above, just go up and to the right. then it must be has path like n + (n-1).

after you know the pattern, let's calculate the path P with that pattern n. it must 25 + (25-1) = 49.

now we know about every path P that we choose randomly, it absolutely has 49 grid points on total of 625 grid points.

then very clear for me that the probability must 49/625.

so, a/b = 49/625 then a + b = 49 + 625 = 674

as you know, I'm not good to write anything like this. and I write this one because it doesn't cost so much time.

and sorry, I don't know so much about math formula, so I often use logical and imagination to solve many problem.

We do have (the number of intersection points in ) / (the number of points in) which can be simply written as P / S There are 25 streets running in each direction, so we have S consists of 625 intersections. On the other hand, P contains other intersections, each one being either one step to the right or one step up from an adjacent intersection in P. There are 25 steps from the top to the bottom street , and 24 from the rightmost to the leftmost one ( or vice versa ) so P = 24 + 25 = 49

The probability is : P /S = a/b = 49/625 , so a+b= 674

To get from the bottom left corner to the top right corner, each path needs to consist of exactly $24$ moves rightwards and exactly $24$ moves upwards. This means each path passes through exactly $49$ intersections. There are thus $49|P| \, (s,p)$ pairs such that $s$ is contained in $p$ . (Here, $|P|$ refers to the number of paths.) $S$ contains $25 \times 25 = 625$ intersections, so there are $625|P| \, (s,p)$ pairs. Since each pair will be chosen with uniform probability, then the probability that $s$ is contained in $p$ would be $\dfrac{49|P|}{625|P|} = \frac{49}{625}$ . Our answer is thus $674$ .

As the assumption states, there are $25*25 = 625$ intersections in $S$ . Notice that there are always a total of $25 + 24 = 49$ intersections in any paths from the bottom left to the top right corner. ( $25$ ups and $25$ right minus $1$ double counted intersection.)

By first randomly choose any paths in $P$ with a probability of $1$ , only $49$ out of $625$ intersections will be contained in that chosen path. Hence, the probability is just $\frac{49}{625}$ and $a+b = \boxed{674}$ .

The number of points in set $S$ is $625$ , and in each path that in the set $P$ there are $25 \times 2-1=49$ points. So, the probability that the point $s$ is contained in the path $p$ is $\frac{a}{b}=\frac{49}{625}$ . Hence, $a+b=49+625=674$

The question is equivalent to ask the probability that a given point from a set of 625 belongs to a path which crosses 49 points. Hence this probability is 49/625.

There are $\binom{48}{24}$ elements in $P$ , and $25^2 = 625$ elements in $S$ , for a total of $625\cdot\binom{48}{25}$ pairs $(s,p)$ . To find the number of pairs $(s,p)$ for which $p$ contains $s$ , note that each path contains exactly 49 intersection points, so there are $49\cdot\binom{48}{24}$ such pairs. So the probability is $\frac{49\cdot\binom{48}{24}}{625\cdot\binom{48}{24}} = \frac{49}{625},$ and the answer is $49+625 = 674.$

all the paths will will include exactly one square of each diagonal lines going from top left to bottom right corners.
thus the sum of the paths passing through the points in a diagonal = total no of paths.
there will be 25 * 2-1 =49 such diagonals.
probability = (sum of number of paths through each point/total number of pats)/total no of paths.
=(49*total no. of paths/total no. of paths)/625.
=49/625.
thus answer = 49+625 = 674

To go from the bottom right to top left we need to take 49 steps (anyhow)

Total number of points is 625

Probablity is 49/625

Answer:: 674

Let's consider the $25 \times 25$ gird in the $xy$ plane. Starting point has coordinate $(0,0)$ while the ending point $(25,25)$ . The number of all possible paths is

$\displaystyle |P|=\binom{24+24}{24}$

and all points on the grid are

$|S|=25^{2}$ .

Let's consider a generic point $s=(h,k), \hspace{8pt} 0 \leq h \leq 24, \hspace{8pt} 0 \leq k \leq 24$ . The number of paths $P[(h,k)]$ passing through $(h,k)$ is

$\displaystyle |P(s)|=\binom{h+k}{h} \binom{24-h+24-k}{24-h}$ .

So, the probability of $(s,p(s))$ is

$\displaystyle \mathbb{P}[(s,p(s))]=\bigg[25^2 \binom{48}{24}\bigg]^{-1} \sum_{h=0}^{24} \sum_{k=0}^{24} \binom{h+k}{h} \binom{24-h+24-k}{24-h}=\frac{49}{625}=\frac{a}{b}$

Eventually

$a+b=\boxed{674}$

The probability of choosing a point s at random from a 25 x 25 grid can be expressed as 1/625. The probability of choosing a path p at random from P can be expressed as 1/49. Dividing 1/625 by 1/49 shows the probability that the point we randomly chose from S is on the set of paths P. This leads to 49/625. a=49 b=625 a+b=674

let it be a N X N grid. clearly there are N^2 intersections. Now, in order to travel from the bottom left corner to the top right corner, we have to travel N-1 units to the right and N-1 units upwards. In moving one unit right or up , we gain one intersection. thus in the whole process, intersections covered = 2(N-1) +1(the bottom left corner from where we started). = 2N-1 thus probability = 2N-1/N^2 in this case, N=25. so probability = 49/625. 49+625=674.

For every path there are 49 points that lie in the path

Therefore probability=(49 * 50C25)/(625 * 50C25)=49/625