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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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    This is a repeat post: http://math.stackexchange.com/questions/84864/probability-urn-question2011-11-23

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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 $\frac{n-m}{n}$.

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    @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.2011-11-23