Question

An urn has n white and m black balls. Balls are randomly withdrawn, without replacement, until a total of $k,k\dots n$white balls have been withdrawn. The random variable $X$equal to the total number of balls that are withdrawn is said to be a negative hypergeometric random variable.

(a) Explain how such a random variable differs from a negative binomial random variable.

(b) Find $P\{X=r\}$.

Hint for (b): In order for $X=r$ to happen, what must be the results of the first$r-1$ withdrawals?

Short answer

$\left(a\right)$Trials are not independent.

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

Step-by-step solution

Given Information (Part a)

Given:

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

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

$X=$Total number of balls withdrawn

$X$has a negative hypergeometric allocation.

Calculation (Part a)

$\left(a\right)$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).

Step 3:Final answer (Part a)

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

Given Information (Part b)

Given:

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

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

$X=$Total number of balls withdrawn

$X$has a negative hypergeometric allocation.

Calculation (Part b)

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 )

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

Final answer (Part b)

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