# Probability selection with replacement (balls in an urn)

• Sep 20th 2012, 08:06 PM
JGWENTWORTH
Probability selection with replacement (balls in an urn)
Hello,

I have a question regarding this subject. If we have replacement, technically the odds of pulling one colored ball remains he same form draw to draw, thus they're independent events, right? Here is a specific example.

There is an urn with 80 blue balls, 60 red balls, and 60 green balls.

If I draw from this urn twice, with replacement, what is the probability that at least one ball is blue given that at least one ball is either green or red?

I go back and forth on this... since it is replacement, part of me says that there will always be a 40% chance that any ball pulled is blue. However, if we have the scenario where there is two balls, and we know that for sure one of the balls chosen will be red or green (60%), would I multiply 80/200 * 120/200?

What is the probability that at least one ball is blue given that the SECOND ball is either green or red?

Does it matter that we know that the second ball is either green or red or is this question to be approached the same way we would answer the reverse - what is the probability that at least one ball is blue given that the FIRST ball is either green or red.... same question? My gut says that this is the same answer as above.

Any insight that could help me get this to click would be greatly appreciated. Thank You.
• Sep 20th 2012, 09:25 PM
MaxJasper
Re: Probability selection with replacement (balls in an urn)
Quote:

Originally Posted by JGWENTWORTH

There is an urn with 80 blue balls, 60 red balls, and 60 green balls. If I draw from this urn twice, with replacement, what is the probability that at least one ball is blue given that at least one ball is either green or red?

Your restrictions give only two drawing possibilities: $\displaystyle \{g,b\}$ and $\displaystyle \{r,b\}$

Combined probabilities would be:

$\displaystyle P\{g,b\}+P\{r,b\}=P(g=1|60,200)*P(b=1|80,200)+P(r= 1|60,200)*P(b=1|80,200)=\frac{6}{25}$
• Sep 23rd 2012, 12:39 PM
JGWENTWORTH
Re: Probability selection with replacement (balls in an urn)
Thanks MaxJasper, in short on this one we're 4/10*6/10. This was the approach that I took. If we had an urn with 20 total balls, 10 yellow and 10 orange, in it and selected 2 balls with replacement, what is the probability that at least one of the balls is orange given that at least one is yellow? Would I take the same approach here? 1/2 * 1/2 = 1/4 ? I see 2/3 being a possibility too as we're looking at OO, YY, OY, YO as possibilities and since we know at least one is yellow, that eliminates OO, leaving us 3 options, YY, OY, and YO... and if at least one of the balls is orange, we have 2 of these 3 possibilities remaining... am I over thinking this? Thanks again!
• Sep 23rd 2012, 06:43 PM
MaxJasper
Re: Probability selection with replacement (balls in an urn)
Again your restrictions specified as "at least" results in only {Y,O} and remember that {Y,O}={O,Y} and so you end up with only {Y,O} no matter what order of drawing the balls. You don't have combinations: {O,O}, {Y,Y} because "at least" is violated!
• Sep 23rd 2012, 07:05 PM
JGWENTWORTH
Re: Probability selection with replacement (balls in an urn)
This being said, the solution for the second question would be 1/2*1/2=0.25 ?