Question

a bag contains 6 red marbles, 4 blue marbles, 7 green marbles, and 3 yellow marbles. a marble is drawn from the bag and is not replaced. Then a second marble is drawn. What is the probability that both marbles were green.

Answer 1

Answer:

7/20 * 6/19 = 21/190 because their are 20 marbles and 7 green then their are one less marbles so only 6 green and 19 total.

brainliest plz :D

Answer 2

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷ This problem involves dependent probability, because you DO NOT replace the marble.

First, you need to find the probability of getting a green marble on the first draw:

7/20 because there are 7 green marbles and 20 marbles total.

Next, you need to find the probability of getting a green marble on the second draw:

6/19 because you took out one green marble there are 6 left and 19 total marbles left.

Then, to get your final probability, multiply the two fractions together:

7/20*6/19=21/190

Final answer:

The probability of getting two green marbles without replacing them is 21/190.

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➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

TROLLER