The phrase

__is a huge clue on this problem.__

**without replacement**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

*we multiply each draw probability together:*

__independent__,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**