# Probability of selecting two different colored balls

##### New Member
Given an urn with x red balls, and y green balls, thoroughly mixed, what is the probability of choosing two colors in the first two picks, when the first pick is put back before the second pick?

#### Dason

You can get this one. There are only two ways this could happen: 1) You could pick a red and then pick a green. Or you could 2) pick a green and then pick a red.

What is the probability of each of these?

#### hlsmith

##### Less is more. Stay pure. Stay poor.
x and y are variables (placeholders)

There are two ways it could go, however if there are an equal number of balls per group, x and y are interchangeable when plugging in data.

All just minutiae, perhaps.

#### Dason

There are two ways it could go, however if there are an equal number of balls per group, x and y are interchangeable when plugging in data.

All just minutiae, perhaps.
I'm not sure what you're getting at but the answer is the same regardless of if x and y are the same.

#### hlsmith

##### Less is more. Stay pure. Stay poor.
Yeah that is correct, then why does it matter that there are two scenarios in this case?

Just making small talk, but make sure you stay away from the public library that crap can literally live for a while. Pun intended

##### New Member
I want a generalized formula that will work with every possible positive integer x and y.
I want a formula that does not use any math notation beyond standard algebra.
I will use the formula to compute the answer.
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I have a bag with 389 black balls, and 2236 white balls. If I randomly draw pairs out of the bag, what number of black/white pairs will I get? What number of black/black pairs will I get? What number of white/white pairs will I get?

#### Dason

What have you tried so far?

##### New Member
Dason thank you for the multiple replies, but I do not understand this forum, nor your replies. Why am I getting questions? Why does no one post the answer?

In seven replies I have not seen even one math formula. What is with that? What I am looking for is a formula, that I expect will contain fewer characters than your last post does. The formula cannot possibly be modified by anything I have tried, and someone who knows the formula does not need to know what I have tried in order to be able to post it. I have tried looking for examples on the web. Going to a dozen web sites. If you search the way I do you will find a plethora of problem sets from university courses, but no solutions. I might be searching with bad search terms. If you search and find it, then send me your search string. I am a 60 year old man with a doctorate in biophysics. I just want the formula.

#### hlsmith

##### Less is more. Stay pure. Stay poor.
I believe you just take x/x+y times y/x+y

We have great probability people here, so we can wait to see if they chime in. Dason was just indirectly referring to posting ettique, of having poster clarify what they have tried. This clears up students just trying to get answers and us telling people to try approaches they have already tried.

#### Dason

I believe you just take x/x+y times y/x+y
Which would give you the probability of a red and then a green. You still need to account for the probability of a green and then a red. (This is what I was talking about before). So take 2*x*y/(x+y)^2 to get the overall probability of choosing different colors in this manner.

Dason was just indirectly referring to posting ettique, of having poster clarify what they have tried.
Pretty much. I don't tend to answer too many questions directly and instead try to help guide people to the solution. It's a better way to learn...

#### BGM

##### TS Contributor
For more general information you may try to search Binomial distribution on the web.

So the answer in the form of Binomial pmf will be like

$$\binom {2} {1} \left(\frac {x} {x + y}\right)^1 \left(\frac {y} {x + y}\right)^{2-1}$$

As mentioned above, it will be much better to let you think why you have an answer like this before we directly provide it. And actually the rationale behind is not that difficult.