91Ó°ÊÓ

Given:

An urn contains $n$white balls and $m$black balls

Balls are removed without replacement until $k$white balls are drawn$(k≤n)$

$X=$Total number of balls withdrawn

$X$has a negative hypergeometric allocation.

$\left(a\right)$Formula Negative binomial distribution:

$P(X=x)=\left(\begin{array}{l}x-1\\ r-1\end{array}\right)p^r(1-p{)}^{x-r}$

The negative binomial allocation is the probability distribution of a variable that counts the number of losses required to get the $r$the success (with independent trials and constant probability of success).

Since the balls are drawn without a substitute, the trials are not independent and thus it is not suitable to utilize the negative binomial distribution (as this distribution needs independent trials).

Trials are not independent (while independent trials is a property of the negative binomial random variable)

Given:

An urn contains $n$white balls and $m$black balls

Balls are removed without replacement until $k$white balls are drawn $(k≤n)$

$X=$Total number of balls withdrawn

$X$has a negative hypergeometric allocation.

Formula hypergeometric probability:

$P(Y=x)=\frac{\left(\begin{array}{l}r\\ x\end{array}\right)\left(\begin{array}{l}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}$

When $X=r$, then we select the $k^{th}$white ball on the $r^{th}$draw, while we then also drew $k-1$white balls and $(r-1)-(k-1)=r-k$black balls on the first $r-1$draws.

By the definition of hypergeometric, there are $P(Y=k-1)=\frac{\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(\begin{array}{c}m\\ rk\end{array}\right)}{\left(\begin{array}{c}n+m\\ r-1\end{array}\right)}$ways to then make the first $r-1$draws.

Since there are then $n-k+1$white balls and $m-(r-k)$black balls remaining after the first $r-1$draws, the probability of drawing a white ball on the $r^{th}$draw is $\frac{n-k+1}{n-k+1+m-(r-k)}=\frac{n-k+1}{n+m-r+1}$.

$P(X=r)=P(Y=k-1)P$( White on $r^{th}$draw )

$=\frac{\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(\begin{array}{c}m\\ r-k\end{array}\right)}{\left(\begin{array}{c}n+m\\ r-1\end{array}\right)}·\frac{n-k+1}{n+m-r+1}$

We have found to be$P(X=r)=\frac{\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(\begin{array}{c}m\\ r-k\end{array}\right)}{\left(\begin{array}{c}n+m\\ r-1\end{array}\right)}·\frac{n-k+1}{n+m-r+1}$.