If 6 red and 4 green marbles are in a bag and you draw two marbles without replacement, what is the probability both are red?

If 6 red and 4 green marbles are in a bag and you draw two marbles without replacement, what is the probability both are red?

• A. 2/5
• B. 1/2
• C. 1/3
• D. 4/9

Answer: (C)

The main idea is that drawing without replacement changes the mix of marbles each time, so you multiply the chances of each successive favorable event. First, the chance the first marble is red is 6 out of 10, or 6/10. If that occurs, there are now 5 red marbles left out of 9 total marbles, so the chance the second one is red is 5/9. Multiply these probabilities: (6/10) × (5/9) = 30/90 = 1/3. An equivalent way to see it is to count combinations: the number of ways to pick two red marbles from the six available is C(6,2) = 15, and the number of ways to pick any two marbles from ten is C(10,2) = 45, so 15/45 = 1/3. So the probability of drawing two red marbles is 1/3 (about 0.333).

The main idea is that drawing without replacement changes the mix of marbles each time, so you multiply the chances of each successive favorable event.

First, the chance the first marble is red is 6 out of 10, or 6/10. If that occurs, there are now 5 red marbles left out of 9 total marbles, so the chance the second one is red is 5/9. Multiply these probabilities: (6/10) × (5/9) = 30/90 = 1/3.

An equivalent way to see it is to count combinations: the number of ways to pick two red marbles from the six available is C(6,2) = 15, and the number of ways to pick any two marbles from ten is C(10,2) = 45, so 15/45 = 1/3.

So the probability of drawing two red marbles is 1/3 (about 0.333).