Answer should be Total no of cases in which they are meeting/ Total no of cases
Total no of cases are : [(5+5)!/5!5!]^2 which is equal to [10!/5!5!]^2 Now for the case of meeting: let number of horizontal steps moved by A is h and no of vertical steps moved by A is v So, no of horizontal steps moved by B is 5-h and vertical steps moved by B is 5-v Also h+v should be equal to 5 as h+v should be equal to 10-(h+v).i.e total steps moved by A and B should be equal as they are moving with same pace
So,numerator should be equal to [summation([(h+v)!]^4/[h!]^4[v!]^4)] from h=0 to h=5 On calculating answer should be 253/756 Why people are calculating answers as 63/256?
Some hints:
Draw a figure! At which lattice points can A, resp., B be after $1$, $2$, $\ldots$ steps? At which lattice points can they meet, if at all? For each of these points, what is the probability that both get there (necessarily at the same time)?
Some more: A hits the diagonal $j+k=5$ in exactly one point, after exactly five steps. So far he had $2^5=32$ choices. The probability $p_j$ that he thereby arrives at the exact point $(j,5-j)$ is given by $$p_j={1\over 2^5}{5\choose j}\qquad(0\leq j\leq5)\ .$$ (Think of Pascal's triangle!). – Same thing for B. The probability $p$ that they hit the same point on this diagonal therefore is $$p=\sum_{j=0}^5 p_j^2=\ldots={63\over256}\ .$$