math probability

An urn contains $n$ balls, all different colors. A person draws a ball randomly, records the color before replacing it. The person must record all colors in order to receive a large prize. If $m$ colors have been recorded so far, what is the probability that it will take exactly $x$ draws to get a new color?

By Bernoulli trial, the probability of a success would be : $(N-M)/M$. Probability of a failure is $M/N$. now the probability that it will take $x$ trials to get a success would be, =>

$$\frac{N-M}{M} \left(\frac{M}{N}\right)^{1-X}$$ => am i correct?

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What do $N$, $M$ stand for and what is the relationship between them and $n$, $m$? –  tards Nov 23 '11 at 15:41
This is related to a well-known problem called the Coupon Collector problem. Have you looked that up? –  Dimitrije Kostic Nov 23 '11 at 15:41
Please unifiy uppercase and lowercase letters. Also, check the denominator in your probability of a success expression. –  leonbloy Nov 23 '11 at 15:43
@DimitrijeKostic: " The person must record all colors in order to receive a large prize. " sounds as the Coupon Collector problem. But the real question (what follows) is not. –  leonbloy Nov 23 '11 at 15:43
@leonbloy I can see that the question being asked isn't precisely the Coupon Collector problem. The question contained no reference to "the Coupon Collector" problem so I said, correctly, that this question is "related." –  Dimitrije Kostic Nov 23 '11 at 16:14

I will interpret "it will take $x$ trials" to mean exactly $x$. So we want the probability of $x-1$ consecutive "failures" followed by a "success." The probability of "failure" is $\frac{m}{n}$. The probability of "success" is $1-\frac{m}{n}$, or equivalently $\frac{n-m}{n}$. The required probability is $$\left(\frac{m}{n}\right)^{x-1} \left(\frac{n-m}{n}\right).$$
Comment: The probability of failure in any one trial is $\frac{m}{n}$ because there are $n$ colours available, all equally likely, and we already have $m$ colours, so will not be happy if we get one of those.
The term $\left(\frac{m}{n}\right)^{x-1}$ in the answer gives the probability of $x-1$ successive failures.
The expression given in the post is related to the correct expression, but differs from it in a couple of ways. It has $1-x$ instead of $x-1$, and $\frac{n-m}{m}$ instead of $\frac{n-m}{n}$.
@user19974: If the probability of failure on any trial is $q$, then probability of fail, fail is $q^2$, the probability of fail, fail, fail is $q^3$, in general the probability of $k$ failures in a row is $q^k$. In our case, we want $x-1$ failures in a row (then a success). The probability of $x-1$ failures in a row is $q^{x-1}$. Remember, $x$ is a positive integer. For most such $x$, the exponent you used, namely $1-x$, will be negative. Since $q$, which is $m/n$, is typically less than $1$, $(m/n)^{1-x}$ will be bigger than $1$, definitely not a probability. –  André Nicolas Nov 23 '11 at 17:28