A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a black and then a red marble out of the bag without replacing the black marble before drawing the red marble?

Ok so there is a total of 12 marbles.

drawing a black is a probability of 3/12. drawing a red is a probability of 4:12

So first it is 3/12 and then because three are drawn instead of out of 12 it is out of 9 for red, so 4/9

3/12 x 4/9

the answer is 1/9

There are a total of 12 marbles so you will start with a denominator of 12.

The P (black) = 3/12 = 1/4

Since the marble was NOT replaced there are now only 11 marbles in the bag.

The denominator will now be 11.

The P (red) = 4/11

Now we multiply these two probabilities (1/4) (4/11) = 1/11 chance of drawing a black and then a red.