find the probability of picking a red marble and a green marble when 2 marbles are picked (without replacement)from a bag containing 6 red and 6 green marbles.}

p = 6/11.

So we have a bag that contains 6 red marbles and 6 green marbles.

Then the total number of marbles that are in that bag is:

6 + 6  = 12

There are 12 marbles in the bag, and we assume that all marbles have the same probability of being randomly drawn.

Now we draw two marbles, we want to find the probability that one is red and the other is green.

The first marble that we draw does not matter, as we just want the second marble to be of the other color.

So, suppose we draw a green one in the first attempt.

Then in the second draw, we need to get a red one.

The probability of drawing a red one will be equal to the quotient between the red marbles in the bag (6) and the total number of marbles in the bag (12 - 1 = 11, because one green marble was drawn already)

Then the probability is:

p = 6/11.

Notice that would be the exact same case if the first marble was red.

Then we can conclude that the probability of getting two marbles of different colors is:

p = 6/11.

i am thinking c. sorry if im wrong .

step-by-step explanation:

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