I'm having trouble with the following problem:
A box contains $n$ identical balls numbered $1$ through $n$. Suppose $k$ balls
are drawn in succession.(a) What is the probability that $m$ is the
largest number drawn?(b) What is the probability that the largest
number drawn is less than or equal to $m$?
The results are (according to the solutions manual), respectively:
$$\frac{m-1 \choose k-1 }{n \choose k }$$
$$\frac{m \choose k }{n \choose k }$$
My understanding of the problem is as follows:
There are $n$ balls in total, $k$ are drawn, so there are $k$ random numbers from $1$ to $n$. I'm asked to find if the greatest number between all the ones that were drawn ($k$), equals $m$ (for part a) and $\le m$ (for part b).
I get that the binomial coefficient in the denominator are all the different ways to arrange in a group of $k$ the $n$ balls. But I have no clue on how to find the numerator.
Thanks in advance.
Best Answer
(a) The largest number drawn is $m$ precisely if $m$ is drawn, and all the other $k-1$ balls drawn are $\le m-1$. There are $\binom{m-1}{k-1}$ ways to choose $k-1$ balls from the balls numbered $1$ to $m-1$.
(b) The probability that the balls are all $\le m$ is simpler. There are $m$ such balls, and therefore $\binom{m}{k}$ ways to choose $k$ of them.
Remark: The answer to (b) can be used to find an alternate form of the answer to (a). The number of ways to have $m$ be largest can, by (b), be written as $\binom{m}{k}-\binom{m-1}{k}$.