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? a.9/25 b.2/15 c.4/25 d.1/3

There are 10 chips altogether. 4 of them are white.
4/10 is the chance of lifting out a white chip
There are 3 of them left and 9 chips altogehter.
4/10 * 3/9
12/90
4/30
2/15

Comment
(my edit) it is not that 2/15 is wrong (although it is not entirely right).
1/3 is the correct answer if you assume that what happened during the first draw has nothing to do with what will happen on the second. It is like saying if you throw 11 heads in a row with a fair coin, what are the chances of throwing a heads on the 12th throw? The answer is 1/2. That is the same kind of question you have asked.

The two of us who have responded have really responded to what are the chances of drawing 2 white chips. The question really does not restrict us in a way that prevents us from saying that. I'll stick with
B <<<< answer
but I think it would be nice if the writer of the question made it clear that 1/3 should be the proper answer. I am glad you came back and posted the right answer. It makes me think. 

The semi right answer is B <<<< 

If my reasoning bothers anybody, I'll reedit again. I'm only leaving it because sometimes a mistake is more instructive than a given answer.

Answer is 2/15

There are initially 4+6=10 chips in the bag, so the probability of the first chip drawn being white is:

4/10=2/5

Having drawn a white chip there are then 3 white chips left in the bag out of a total 3+6=9 chips. So the probability of the second chip also being white is:

3/9=1/3

So the probability of drawing two white chips is:

2/5x1/3=2/15

Pls mark brainliest

1. 5/82.95% is better than 7/8