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.

Let  be the probability that a single toss will come up tails.

Assume Statement 1 alone. 

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 .

The probability of four tails in a row on the altered coing will be , which is less than  by Statement 2. Therefore,

,

The probability of one toss of the coin coming up tails decreased, so the probability of it coming up heads increased.

A similar argument can be used to demonstrate that Statement 2 allows the same conclusion to be drawn.

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 Dick replaced with the joker was a black card. Statement 2, but not Statement 1, provides this information. Note that Statement 1 is irrelevant - whether Dick knew the color of the card does not change any theoretical probability.

In order to calculate probability, we need to know the total number of coins, as well as the number of each type of coin.

In this case, I) tells us how many types of coins there are, and II) gives us a clue about the ratio of 5 euro to 50 cent coins.

The key information we are missing is the amount of 2 euro coins that are in the bag. Without this piece of information neither statement will allow us to answer the question.

Therefore, we do not have enough information to answer this question.