May 26, 2:15:28 PM

A bag contains 7 red marbles, 3 blue marbles and 8 green marbles. If three marbles

are drawn out of the bag, what is the probability, to the nearest 10th of a percent, that

all three marbles drawn will be green?

Answer:

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attempt i out of 2

6.9%

Step-by-step explanation:

7 red marbles, 3 blue marbles and 8 green marbles = 18 marbles

P (green) = green/total = 8/18 = 4/9

Keep marble

7 red marbles, 3 blue marbles and 7 green marbles = 17 marbles

P (green) = green/total = 7/17

Keep marble

7 red marbles, 3 blue marbles and 6 green marbles = 16 marbles

P (green) = green/total = 6/16 = 3/8

P (green, green, green) = 4/9*7/17 * 3/8 = 7/102

To the nearest tenth of a percent

.068627451 = 6.8627451% = 6.9%