I have 3 different balls and I extract one of them three times with replacement.

Which is the probability of extracting each ball one time? What rule did you apply?

Thanks —The preceding unsigned comment was added by Daino3z (talk • contribs) 15:24, July 16, 2012‎ (UTC).

At each extraction, the probability of choosing the given ball is 1/3, and choosing another one is 2/3

To choose the given ball exactly once out of three extractions can occur 3 different ways: Choose it in the first extraction and not in the other two (probability 1/3 times 2/3 times 2/3 = 4/27, multiplying the probabilities because these extractions are independent of each other), or choose it in the second extraction and not in the other two (probability 2/3 times 1/3 times 2/3 = 4/27), or choose it in the third extraction and not in the other two (probability 2/3 times 2/3 times 1/3 = 4/27). So the total probability of one of these three events happening is the sum of their probabilities, 4/27 + 4/27 + 4/27 = 12/27 = 4/9, adding the probabilities because these events are disjoint, ie more than one cannot occur at the same time.

The multiplying part is sometimes called the "and" rule (ie, draw right ball AND then draw wrong ball AND then draw right ball, replace the ANDs with times) and the adding part is called the "or" rule (ie, the correct ball will be in the first draw OR it will be in the second draw OR it will be in the third draw, replace the ORs with add)

That's if I've understood your question right--Acer4666 (talk) 00:16, July 17, 2012 (UTC)