Monique has a box that contains the following marbles: 5 yellow marbles, 3 green marbles, 2 orange marbles. She randomly selects 2 marbles from the box and keeps them. If she randomly selects a third marble from the box, the probability that she selects a green marble will be 2/10.

Answer 1

No, this statement is false. Let's calculate the probability of Monique selecting a green marble in her third draw after she has picked 2 marbles and kept them:

Total marbles in the box initially = 5 (yellow) + 3 (green) + 2 (orange) = 10

There are 3 possibilities for her first two selections:

1) She selects 2 yellow marbles (probability 5/10 * 4/9): The box will then contain 1 green and 2 orange marbles, and the probability of selecting a green marble will be 1/3.

2) She selects 1 yellow and 1 green marble or 1 green and 1 yellow marble (probability 5/10 * 3/9 + 3/10 * 5/9): The box will then contain 2 green and 2 orange marbles, and the probability of selecting a green marble will be 2/4 = 1/2.

3) She selects 2 green marbles (probability 3/10 * 2/9): The box will then contain 1 green and 2 orange marbles, and the probability of selecting a green marble will be 1/3.

Now let's calculate the overall probability of Monique selecting a green marble in her third draw using these probabilities:

Overall_probability = P(draw_green|GG) * P(GG) + P(draw_green|GY) * P(GY) + P(draw_green|YY) * P(YY)

Overall_probability = (1/3) * (3/10 * 2/9) + (1/2) * (5/10 * 3/9 + 3/10 * 5/9) + (1/3) * (5/10 * 4/9)

Overall_probability = 8/27 + 1/2 + 1/9 ≈ 0.481

This overall probability (48.1%) of Monique selecting a green marble in her third draw is different from 2/10 (20%); hence, the statement is false.