There are $N$ coupons numbered $1,2,\dots,N$ in a box. If $n$ coupons are taken at random, what is the probability that $M$ will be the highest number on a ticket if the drawings are made with replacements?
The answer can be $\frac{M^n-(M-1)^n}{N^n}$. Because, $n$ tickets can be drawn with number at most $M$ is $M^n$. Now subtracting the cases where the number is at most $M-1$, we get the above answer.
But why isn't the following answer correct? $\frac{(M-1)^{n-1}}{N^n}$
We have to draw ticket no. $M$ necessarily. So now we have to draw the other $n-1$ tickets from the $M-1$ tickets with numbers less than $M$, which is $(M-1)^{n-1}$.
- Why isn't this answer correct? Where's actually the mistake? This is the place where I'm stuck right now.
Your mistake is that implicitely you assume that in the remaining $n-1$ draws, you never get $M$. But you can. This makes the computation more complicated. Idea: compute the number of possibilities where the maximum is AT MOST $M$. If this is $f(M)$, then you are looking for $f(M)-f(M-1)$.