# win/lose probability

• Mar 27th 2009, 10:24 AM
antman
win/lose probability
An urn contains five balls, one marked win and four marked lose. You and another player take turns selecting a ball from the urn , one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done

a. with replacement
b. without replacement

My work:
a. (1/5)+(4/5)(1/5)+(4/5)(4/5)(1/5)=5/9
b. (1/5)+(4/5)(3/4)(1/3)+(4/5)(3/4)(2/3)(1/2)=3/5
• Mar 27th 2009, 11:40 AM
running-gag
Quote:

Originally Posted by antman
An urn contains five balls, one marked win and four marked lose. You and another player take turns selecting a ball from the urn , one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done

a. with replacement
b. without replacement

My work:
a. (1/5)+(4/5)(1/5)+(4/5)(4/5)(1/5)=5/9
b. (1/5)+(4/5)(3/4)(1/3)+(4/5)(3/4)(2/3)(1/2)=3/5

Hi

I agree with your results (5/9 and 3/5) even if I do not understand how you manage to find 5/9 for a.

$\frac15 + \frac{4}{5}\:\frac{4}{5}\:\frac{1}{5}+\frac{4}{5}\ :\frac{4}{5}\:\frac{4}{5}\:\frac{4}{5}\:\frac{1}{5 } \cdots = \frac15\:\left(1+\frac{16}{25}+\left(\frac{16}{25} \right)^2+\cdots \right)= \frac15\:\frac{1}{1-\frac{16}{25}}=\frac59$
• Mar 27th 2009, 02:41 PM
Soroban
Hello, antman!

and I think I know how you got $\tfrac{5}{9}$.

Quote:

An urn contains five balls, one marked WIN and four marked LOSE.
You and another player take turns selecting a ball from the urn , one at a time.
The first person to select the WIN ball is the winner.
If you draw first, find the probability that you will win if the sampling is done

. . a. with replacement . . b. without replacement

My work:

a. (1/5) + (4/5)(1/5) + (4/5)(4/5)(1/5) = 5/9 .??

b. (1/5) + (4/5)(3/4)(1/3) +(4/5)(3/4)(2/3)(1/2) = 3/5

For (a), I believe you meant to type: . $P \;=\;\frac{1}{5} + \left(\frac{4}{5}\right)^2\frac{1}{5} + \left(\frac{4}{5}\right)^4\frac{1}{5} + \hdots$

$\text{Then: }\;P \;=\;\frac{1}{5}\underbrace{\bigg[1 + \left(\frac{4}{5}\right)^2 + \left(\frac{4}{5}\right)^4 + \hdots \bigg] }_{\text{geometric series}}$

The geometric series has sum: . $\frac{1}{1-\frac{16}{25}} \:=\:\frac{1}{\frac{9}{25}} \:=\:\frac{25}{9}$

. . Therefore: . $P \;=\;\frac{1}{5}\cdot\frac{25}{9} \;=\;\frac{5}{9}$

If that is the case, you did terrific work on both parts.

. . Nice going!