There are 4 people and 4 marbles (3 white and 1 black) . Each take one marbles. What is the chance that the ball first person takes is black, the second person takes is black... .

Are these all separate events?

prob(first person takes black, others take white):
BWWW

= (1/4)(3/3)(2/2)(1/1) = 1/4

The others would be the same,
note that all cases would have to add up to 1, they do.

Then p(second person takes black)= 3/4* 1/3* 2/2*1/1=1/4

How can i draw this situation?

You answer of 3/4* 1/3* 2/2*1/1 = 1/4 , show this situation

WBWW

So was that correct? Do you know how can i draw this in a probability tree?

To find the probability of each person taking a black marble, we can use the concept of conditional probability.

The probability of the first person drawing a black marble is 1/4, as there is only one black marble out of the four total marbles.

After the first person takes their marble, there are now 3 marbles left, 1 of which is black. So, the probability of the second person drawing a black marble, given that the first person already did, is 1/3.

Similarly, after the second person takes their marble, there are now 2 marbles left, 1 of which is black. So, the probability of the third person drawing a black marble, given that the first two people already did, is 1/2.

Finally, after the third person takes their marble, there is only 1 marble left, which must be black. So, the probability of the fourth person drawing a black marble, given that the first three people already did, is 1.

To find the overall probability of all four people drawing black marbles in succession, we multiply the individual probabilities together:

(1/4) * (1/3) * (1/2) * 1 = 1/24

Therefore, the chance that each person takes a black marble in the given scenario is 1/24.