How to calculate the following probability?

Question

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.

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@user587389: Please check that the two expressions I MathJaxified are correct. — Brian Tung, Aug 26 '18 at 19:38
Thank you. I was in search of such a tool with which I can type mathematical expressions. — user587389, Aug 31 '18 at 10:07
@user587389 Don´t forget to mark questions as answered. Otherwise users are less willing to answer your questions.+ — callculus42, Nov 17 '18 at 20:20
@user587389 Just click on the checkmark ($ \checkmark $) for your preferred answer. — callculus42, Nov 18 '18 at 13:44

score 1 · Accepted Answer · answered Aug 26 '18 at 19:00

1

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 $f(M)-f(M-1)$.

answered Aug 26 '18 at 19:00

A. Pongrácz

7,418

Thank you Sir. Now it's clear to me. – user587389 Aug 31 '18 at 10:08

How to calculate the following probability?

1 Answers1