Ask Question
12 December, 08:22

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?

+4
Answers (2)
  1. 12 December, 08:59
    0
    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.
  2. 12 December, 09:01
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers