# Math Help - A urn contain N white

1. ## 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
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Should i be using geometric distribution

2. 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.

3. Hello, zorro!

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

An urn contains ${\color{red}W}$ white and ${\color{red}B}$ 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 $n$ draws are needed.

b) at least $k$ draws are needed.

Should i be using geometric distribution? . . . . no
Let $W + B \,=\,T$, the total number of balls.

Since the balls are replaced, the probabilities remain constant.

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

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

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

(b) At least $k$ draws.

We have this list of possible outcomes:

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

Our probability is the sum of these probabilities:

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

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

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

Substitute into [1]:

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

4. ## Thanks mite

Thanks mite
Happy and prosperous NEW YEAR
Cheers