# A urn contain N white

• Dec 29th 2009, 09:02 PM
zorro
A urn contain N white
Question : A urn contain N white and M black balls. Balls are randomly selected, one at a time , until a black one is obtained . It we assume that each selected ball is replaced before the next one is drawn, what is the probability that
i) exactly n draws are needed
ii) atleast k draws are needed
----------------------------------------------------------------

Should i be using geometric distribution
• Dec 30th 2009, 12:02 AM
mr fantastic
Quote:

Originally Posted by zorro
Question : A urn contain N white and M black balls. Balls are randomly selected, one at a time , until a black one is obtained . It we assume that each selected ball is replaced before the next one is drawn, what is the probability that
i) exactly n draws are needed
ii) atleast k draws are needed
----------------------------------------------------------------

Should i be using geometric distribution

Yes. p = M/(M + N). Now go and thoroughly review that topic.
• Dec 30th 2009, 05:41 AM
Soroban
Hello, zorro!

Who wrote this problem?
What a confusing set of variables!

Quote:

An urn contains $\displaystyle {\color{red}W}$ white and $\displaystyle {\color{red}B}$ black balls.
Balls are randomly selected, one at a time , until a black one is obtained.
Each selected ball is replaced before the next one is drawn.

Find the probability that:

a) exactly $\displaystyle n$ draws are needed.

b) at least $\displaystyle k$ draws are needed.

Should i be using geometric distribution? . . . . no

Let $\displaystyle W + B \,=\,T$, the total number of balls.

Since the balls are replaced, the probabilities remain constant.

. . $\displaystyle \begin{array}{ccc}P(W) &=& \dfrac{W}{T} \\ \\[-3mm] P(B) &=& \dfrac{B}{T} \end{array}$

(a) A black ball is drawn on the $\displaystyle n^{th}$ draw.
. . .The first $\displaystyle n\!-\!1$ balls are White, the $\displaystyle n^{th}$ is Black.

. . . $\displaystyle P(n\text{ draws}) \;=\;\left(\frac{W}{T}\right)^{n-1}\!\!\left(\frac{B}{T}\right)$

(b) At least $\displaystyle k$ draws.

We have this list of possible outcomes:

. . $\displaystyle \begin{array}{ccc}P(k\text{ draws}) &=& \left(\frac{W}{T}\right)^{k-1}\left(\frac{B}{T}\right) \\ \\[-3mm] P(k\!+\!1\text{ draws}) &=& \left(\frac{W}{T}\right)^k\left(\frac{B}{T}\right) \\ \\[-3mm] P(k\!+\!2\text{ draws}) &=& \left(\frac{W}{T}\right)^{k+1}\left(\frac{B}{T}\ri ght) \\ \\[-3mm] \vdots && \vdots \\\end{array}$

Our probability is the sum of these probabilities:

$\displaystyle P(\text{at least }k\text{ draws}) \;=\;\left(\frac{W}{T}\right)^{k-1}\!\!\left(\frac{B}{T}\right) \;+\; \left(\frac{W}{T}\right)^k\!\!\left(\frac{B}{T}\ri ght) \;+\; \left(\frac{W}{T}\right)^{k+1}\!\!\left(\frac{B}{T }\right) \;+\; \hdots$

. . . . . . . . . . . . $\displaystyle =\;\left(\frac{W}{T}\right)^{k-1}\!\!\left(\frac{B}{T}\right)\underbrace{\bigg[1 + \frac{W}{T} + \left(\frac{W}{T}\right)^2 + \hdots \bigg]}_{\text{geometric series}}$ .[1]

$\displaystyle \text{The sum of the geometric series is: }\:\frac{1}{1-\frac{W}{T}} \:=\:\underbrace{\frac{T}{T-W}}_{\text{This is }B} \;=\;\frac{T}{B}$

Substitute into [1]:

. . $\displaystyle P(\text{at least }k\text{ draws}) \;\;=\;\;\left(\frac{W}{T}\right)^{k-1}\!\!\left(\frac{B}{T}\right)\!\left(\frac{T}{B}\ right) \;\;=\;\;\left(\frac{W}{T}\right)^{k-1}$

• Dec 30th 2009, 08:17 PM
zorro
Thanks mite
Thanks mite
Happy and prosperous NEW YEAR (Dance)(Ninja)
Cheers (Beer)