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.

Since the number of cards in the deck remained unchanged, the probability of a random draw resulting in a black card changed if and only if the number of black cards was changed - that is, if the card Corinne replaced with the joker was a black card. Statement 2 alone, but not Statement 1 alone, answers that question.

Neither statement alone is enough to answer the question, since neither alone yields clues as to the ratio of white marbles to black both before and after the addition.

Assume both statements hold. From Statement 1 alone, the probability of drawing a white marble was  before the addition. The probability of drawing a white marble remains  if and only if  of the marbles added were white. If this had happened, since, by Statement 2, seven of the marbles added were white, then the total number of marbles  added could be calculated by solving for :

However, the number of marbles must be an integer. Therefore, for the conditions of both statements to hold, it is impossible for the probability of drawing a white marble to have remained the same.

Statement 1 yields no helpful information. The only way to roll a sum of "12"  with two dice is by rolling "6" on both; the probability of rolling a "6" on a given can change or remain constant without affecting that of rolling a "1".

Assume Statement 2 alone, and let  the probability of rolling a "1" on the altered die. There is only one way to roll a die so that its sum comes up a "2" - a double "1". Since the probability of rolling a "1" on the fair die is , by the multiplication principle, the probability of rolling a double six with the fair die and the die that Henry altered is . Since this is equal to , we solve for :

Henry's alterations increased the probability of a "6" coming up.

The two statements together are insufficient.

Jenny's and Susie's experiments both yield empirical data, which is not in and of itself sufficient to draw a definite conclusion about the theoretical probability of any outcomes. While their results strongly suggest that the probability of the coin coming up heads has increased, it is entirely possible for a fair coin, or even a coin altered to come up tails more often, to be flipped with these results as well.

The probability that a roll of an unaltered die will yield an outcome of "6" is ; the same holds for an outcome of "5". 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 twice so that the sum of the rolls comes up a "11" - a"5" and a "6", in either order. The probability of rolling a "5" and a "6" on the altered die in that order is , which is also that of the reverse event, so the probability of rolling an "11" is . 

On an unaltered die, the probability of this happening would be . 

By Statement 1, the alterations decreased this probability, so

and 

Without further information, it cannot be determined whether  increased, decreased, or remained the same.  

For example, if  and , 

.

If  and , 

.

Both scenarios fit the conditions of Statement 1, but in one, the probability of rolling a "6" on the die remained the same, and in the other, the probability decreased. Statement 1 gives insufficient information.

Assume Statement 2 alone. There is only one way to roll a die twice so that the sum of the rolls comes up a "12" - a double "6". The probability doing this twice on the altered die is . On an unaltered die, the probability of this happening would be ; however, since, by Statement 1, the alterations decreased this probability, 

, 

Xenia's alterations decreased the probability of rolling a "6" on the die.

The two statements together are insufficient.

Quentin's and Allen's experiments both yield empirical data, which is not in and of itself sufficient to draw a definite conclusion about the theoretical probability of any outcomes. While their results strongly suggest that the probability of the die coming up a "6" remained unchanged (since , the probability of the roll of a fair die resulting in an outcome of "6"), it is entirely possible for an experiment with a biased die to yield these results.

Since the number of cards in the deck remained unchanged, the probability of a random draw resulting in a black card changed if and only if the number of red cards was changed - that is, if the card Roger replaced with the joker was a red card.

Statement 2 alone provides this information. Statement 1 is unhelpful, in that it deals with an empirical result. It is entirely possible for this to happen in an unmodified deck, or even in a deck with more black cards than red.