there is a square with $60$ equal blocks.
If a mosquito(bug)is set to fly starting at block $1$, it is equally likely to fly to other blocks. what is the probability after $n$ flies, the mosquito is at the $60\text{th}$ block (diagonally opposite to block $1$). form the difference equation.
I am trying to use law of probability to solve it but cannot form the equation.can anybody here can help. Thank you.
We assume that the fly can be at any one of $60$ positions, and that in any flight, it chooses at random one of the $59$ other positions to go to.
Say it starts at position $A$. We want to find the probability that after $n$ flights it ends up at $B$, where $B$ is a particular position other than $A$.
Let this probability be $p_n$. We have $p_0=0$. The probability that after $n$ flights it ends up at $B$ is the probability it is at a position other than $B$ after $n-1$ flights, and goes to $B$ on the $n$-th. Thus $$p_n=(1-p_{n-1})\cdot\frac{1}{59}.\tag{1}$$ There are various methods to solve linear recurrences. One way is to look first at the *homogeneous recurrence $$p_n=-\frac{p_{n-1}}{59},$$ which has the general solution $p_n=a\left(-\frac{1}{59}\right)^n$ for some constant $a$. Then we look for a particular solution of (1). Here we look for a constant solution $b$. So we want $b=(1-b)\cdot\frac{1}{59}$, which gives $b=\frac{1}{60}$. The general solution of the recurrence (1) is then $$p_n=a\left(-\frac{1}{59}\right)^n.$$ Choose $a$ to meet the initial condition $p_0=0$. We get $a=-\frac{1}{60}$, and therefore $$p_n=\frac{1}{60}\left(1-\frac{(-1)^n}{59^n}\right).$$
Remark: There are many other approaches to the solution.