In order to calculate the probability the question asks for, we need to know the number of each color of marble.

I) Gives us the number of greens.

II) Gives us clues which will allow us to find the number of reds and yellows.

We need both statements to answer this question.

To find the desired probability we need to know which color cards were removed.

I) Tells us the number of black cards removed.

II) Tells us the number of red cards removed.

At first glance, we need both statements, but if we think for a minute, if we know the total number of cards removed and the number of one color removed, we can find the number of the color... So either statement works.

The statements together do not provide a definitive answer. The question addresses theoretical probability, but the two statements together provide empirical results. While the two experiments strongly suggest that Holly removed mostly red cards, it is entirely possible for these results to happen with an evenly distributed deck, or even one which has all of its red cards left.

The probability of a single toss of a fair coin coming up heads (or tails) is ; the probability of five such outcomes in a row is .

Assume Statement 1 alone. Let  be the probability that a single toss of the coin will come up heads. The probability of five such outcomes in a row will be , which is greater than  by Statement 1. Therefore,

,

The probability of one toss of the coin coming up heads increased.

If Statement 2 alone is assumed, a similar argument shows that the probability of one toss of the coin coming up tails decreased - which, of course, is the equivalent outcome.

Statement 1 alone only gives the probability that any of three rolls, "1", "3", or "5", comes up, but it does not give us the probability of a "2" or a "4". The reverse holds for Statement 2. Neither alone gives us the complete picture.

From the two statements together, it can be deduced that the probability of the die not coming up a "6" is the sum , so the probability of the die coming up a "6" is . Since the probability of the die coming up "6" was  before it was altered, the alteration increased this probability.

From either statement, the length of one edge of a cube can be determined:

If the surface area is 150, then 

If the volume is 125, then 

Either way, since the length of each edge is known to be 5, the area of each of , , and  can be calculated by multiplying one half by the product of the legs - that is, one half by the square of 5. 

Also, since the three triangles are all right triangles with the same leg lengths, by the Side-Angle-Side Theorem, they are congruent, and their diagonals are as well - this makes  equilateral. Also, since each triangle is a right isosceles triangle, by the 45-45-90 Theorem, each hypotenuse can be calculated by multiplying 5 by . Therefore, the sidelength of  can be calculated, and its area can be determined. 

Therefore, each statement alone is enough to yield the area of each face - and the total surface area.

From Statement 1 alone, the probability of drawing a red number from the box before the addition of more marbles can be calculated - it is  - but no clue as to the number of red or white marbles in the box after the addition is provided. Statement 2 alone gives no clue about the numbers of marbles before or after the addition.

The two statements together provide sufficient information. If half the marbles in the box before the addition are red, as given in Statement 1, and half the marbles added are red - half the marbles are white, from Statement 2, and no other colors were added - then half the marbles in the box after the addition are red, and the probability remains  .

We show that the two statements together provide insufficient information by looking at two scenarios.

Case 1: Andrea's alterations increased the probability of each individual even outcome. 

As a result of this, the total probability of any even outcome happening must increase, so that of an odd outcome must decrease.

Case 2: Andrea's alterations make the probabilities of the even outcomes as follows:

In an unaltered die, the probability of a roll resulting in a "2" or a "4" is . In this scenario, Andrea's alteration results in the probability if this event being

 .

This scenario satisfies the conditions of Statement 1. By a similar argument, it satisfies the conditions of Statement 2 also.

The probability that the roll will come up even is the sum of these probabilities:

The total probability of any even outcome happening must decrease, so that of an odd outcome must increase.

Let  and  the probabilities of rolling a "5" or a "6", respectively, on the altered die.

Assume Statement 1 alone. There is only one way to roll a die so that its sum comes up a "12" - a double "6". Since the probability of rolling a "6" on the fair die is , by the multiplication principle, the probability of rolling a double six with the fair die and the die that Violet altered is . Since this is equal to , we solve for :

This is the probability of an unaltered die coming up a "6", so Violet's alterations did not affect the probability of a "6" coming up.

Assume Statement 2 alone. The only way to roll an "11" with two dice is to roll a "5" and a "6" - however, either number can be rolled on the altered die. The probability that a "5" is rolled in the altered die and a "6" is rolled on the fair die is ; the probability of the reverse outcome is . The total probability of rolling an "11" is , so the equation is

and

Since the probability of rolling either a "5" or a "6" on a fair die is , this probability has decreased, but without further information, we cannot determine which probability has decreased - that of rolling a "5", a "6", or both.

Assume both statements are true. Let  and  be the number of green and total marbles added, and  and  be the number of green and total marbles in the box after the addition. 

From Statement 1, there were twice as many blue marbles as green after the addition, so green marbles comprised one-third of the marbles; therefore, the probability of drawing a green marble after the addition was .

The probability of drawing a green marble from the box before the addition is  

.

From Statement 1, , and from Statement 2, , so the probability of drawing a green marble before the addition was

.

However, since no clue is given as to how many blue marbles were added - that is, the value of  - it cannot be determined whether this is greater than, equal to, or less than . Consequently, whether the addition of the marbles increased, decreased, or left unchanged the probability of drawing a green one cannot be determined.