Suppose a bag contains 4 white chips and 6 black chips. what is the probability of randomly choosing a white chip, not replacing it, and then randomly choosing another white chip?

There are 4+6=10 chips in the bag. 4 of them are white. The probability that the first chip chosen will be white is 4/10=2/5.
After that, there is one white chip less in the bag, so there are 9 chips in the bag, 3 of them are white. The probability that the second chip chosen will be white is 3/9=1/3.

Multiply the probabilities to find the probability that the first chip will be white and the second chip will be white:


The probability of choosing a white chip and then another white chip is 2/15.

--The probability of choosing a white chip on the first draw is 4/10 .

-- If you already have one white chip, then the probability of choosing
another one on the second draw is 3/9.

--  So the probability of both is  (4/10) x (3/9) = 12/90 = (13 and 1/3) percent .

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.

What exactly is your question?