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 drawn were green? . Option A: 49/380. Option B: 49/400. Option C: 21/190. Option D: 21/200.

Answer 1

$\frac{7}{6 + 4 +7 +3}$ x $\frac{7 - 1 }{ 6 + 4 + 7 + 3 - 1}$

= 21/190

So the answer is option C.

hope this helps

Answer 2

20 total marbles....7 are green

probability on first draw is 7/20...and since the marble was not replaced, the probability on the second draw is 6/19. And since they cant happen at the same time, we multiply

7/20 * 6/19 = 42/380 which reduces to 21/190