Ask Question
17 August, 04:58

10 marbles. 4 are green, 2 are red, and 4 are blue. choosing a marble at random, and without putting it back, chooses another one at random. what is the probability that both marbles he chooses are blue.

+3
Answers (1)
  1. 17 August, 08:24
    0
    12/90 = 6/45

    Step-by-step explanation:

    In the first draw the probability will be 4/10 because there are 4 blue marbles and 10 total marbles. In the second draw, since they did not put the marble back, the probability will be out of 9 marbles. There is also only 3 marbles, so the probability will be 3/9. Since you are calculating the probability of both combined, the answer will be 4/10 * 3/9 = 12/90 which simplifies to 6/45
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “10 marbles. 4 are green, 2 are red, and 4 are blue. choosing a marble at random, and without putting it back, chooses another one at ...” 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