A bag contains 2 red marbles, 3 green marbles, and 4 blue marbles.

If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be green and the second will be red?

Question

Yael

Mathematics

5 November, 13:39

A bag contains 2 red marbles, 3 green marbles, and 4 blue marbles.

If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be green and the second will be red?

+5

Answers (2)

Know the Answer?

Taryn · Answer 1

0.083333333333 or 8.3%

Step-by-step explanation:

3/9x2/8

The second fractions has a demoninator of 8 bc u already took one from the bag and didnt return it

Berlynn · Answer 2

1/12.

Step-by-step explanation:

There are a total of 9 marbles so

Prob (drawing first a green) = 3/9.

After drawing the green there are 8 marbles left in the bag, so:

Prob (drawing a red) = 2/8.

The required probability = 3/9 * 2/8

= 1/3 * 1/4

= 1/12.

A bag contains 2 red marbles, 3 green marbles, and 4 blue marbles.If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be green and the second will be red?

A bag contains 2 red marbles, 3 green marbles, and 4 blue marbles.

If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be green and the second will be red?