A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a b

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?

The phrase without replacement is a huge clue on this problem.

Take each draw and calculate the probability.

Draw 1: P(Drawing a red)
P(Drawing a red) = Total Red marbles n the jar / Total marbles in the jar
P(Drawing a red) = 4/12

4/12 simplifies to 1/3 using a common factor of 4:
P(Drawing a red) = 1/3

Draw 2: P(Drawing a black)
P(Drawing a black) = Total Black marbles in the jar / Total marbles in the jar
We drew one red marble already. Without replacement means we do not put it back. Therefore, we have 12 - 1 = 11 marbles left in the jar.
P(Drawing a black) = 3/11

The question asks, what is the the following probability:
P(Drawing a Red, Drawing a Black)

Because each draw is independent, we multiply each draw probability together:
P(Drawing a Red, Black) = P(Drawing a Red) * P(Drawing a Black)
P(Drawing a Red, Black) = 1/3 * 3/11
P(Drawing a Red, Black) = 1/11