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?

Solution:

Given, a bag contains 6 red marbles, 4 blue marbles, 7 green marbles, and 3 yellow marbles.

Total marbles in the bag = 6 + 4 + 7 + 3 = 20.

We have to find the probability of picking two green marbles.

First, marble is drawn from the bag and is not replaced.

So, the probability of drawing a green marble, P(A) = 7/20

Now, the second marble is drawn.

Since the first drawn marble is not replaced.

Total number of marbles in the bag = 19

So, the probability of drawing another green marble, P(B) = 6/19

Now, the probability of drawing two green marbles = P(A) × P(B)

= (7/20) × (6/19)

= 42/380

= 0.1105

Therefore, the probability of drawing two green marbles is 0.1105

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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?

Summary:

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. The probability that both marbles were green is 0.1105