Probability and simultaneous sampling/selecting

[kakab00]

I was just wondering, how do you calculate the probability of something if you draw a lot of them together at the same time?

For example: there are 10 red balls, 10 blue balls and 10 green balls. You put your hands in and draw out 3 balls at the same time. How would you calculate the probability of getting 2 balls of the same color and 1 ball of different color at the same time?

I know how to calculate if they're taken out slowly 1by1 with or without replacement, but I have no clue how to do this.

 

[chroot]

What is the difference between taking three balls out one at a time, without replacement, and three balls at once? None that I can see.

- Warren

 

[kakab00]

What is the difference between taking three balls out one at a time, without replacement, and three balls at once? None that I can see.

- Warren

I just thought if you took out 3 balls at once there would be an equal chance to get each ball? For example if you wanted 3 balls of the same color then it would be 10/30 x 9/29 x 8/28 if you took them out 1 by 1 without replacement. But if you took 3 balls at once wouldn't it be something like 10/30 x 9/30 x 8/30 or 10/30 x 10/30 x 10/30?

 

[chroot]

Well, let's clarify this just a bit more.

Say you pull out three balls at once, like red, red, and green. Certainly you consider any such handful of two reds and one green to be equivalent, no matter if the green one is on the left or the right in your hand.

This is just like pulling three balls out of the bag, one at a time, but not caring about their final order. In other words, if you pull red, red, green, you should consider that the "same event" as pulling red, green, red. This is a permutation. If you add up the probability of all the ways to pull out two reds and a green, one at a time, it will equal the probability of pulling two reds and a green out simultaneously.

Is that the confusion you're having?

- Warren

 

[kakab00]

Well, let's clarify this just a bit more.

Say you pull out three balls at once, like red, red, and green. Certainly you consider any such handful of two reds and one green to be equivalent, no matter if the green one is on the left or the right in your hand.

This is just like pulling three balls out of the bag, one at a time, but not caring about their final order. In other words, if you pull red, red, green, you should consider that the "same event" as pulling red, green, red. This is a permutation. If you add up the probability of all the ways to pull out two reds and a green, one at a time, it will equal the probability of pulling two reds and a green out simultaneously.

Is that the confusion you're having?

- Warren

Yes, that's the confusion I have. So for the example I gave above, let's say you want to get 2 red balls and 1 blue ball, then the probability would be 10/30 x 9/29 x 1/28 x 3! ??

 

[chroot]

Well, I think you made typo (1 instead of 10/28), but yes, that seems correct to me. About 22% of the time, you'll end up with two reds and a blue.

The breakdown looks like this:

P{all same color} = P{all red} + P{all blue} + P{all green} = 10/30 * 9/29 * 8/28 * 3 = 0.08867

The factor of three is there because there are three possible colors.

P{all different colors} = 0/30 * 10/29 * 10/28 * 3! = 0.246305

The factor of 3! is there because it doesn't matter if you pull RGB or GRB or BRG, etc.

P{two of one color and one of another color} = 10/30 * 9/29 * 10/28 * 3 * 6 = 0.665025

One factor of three is there because there are three permutations of two balls and one ball. For example, if you pulled two reds and a blue, you could have any of RRB, RBR, BRR.

The factor of six is in there because there are six possible combinations of two balls of one color and one of another: RRB, BBR, RRG, GGR, BBG, GGB.

- Warren

 

[kakab00]

That cleared a lot of things up for me already , thanks.