The two statements are actually equivalent; if the loaded die comes up any of 2, 3, 4, 5, or 6 with probability , then it will come up 1 with probability , which is exactly the same assertion as Statement 1 makes.

A sum of 2 can only come up one way: a pair of ones. If you know either statement, then you can calculate the probability of tossing a 2 by multiplying , the probability of a die coming up a 1 on the fair die, by , the probability of a 1 coming up on the loaded die.

The only way the sum of the two dice can come up 12 is a pair of sixes, so the only probabilities that count are those of each die coming up 6; Statement 2 is therefore irrelevant to the problem. If you are given Statement 1, however, then you can calculate the probability of tossing a 12 by multiplying:

Box 1 will contain  white marbles; Box 2 will contain  white marbles.

The probability of choosing Box 1 is , and the probability of choosing a white marble from this box is , making the probability of choosing Box 1, then a white marble, .

The probability of choosing Box 2 is , and the probability of choosing a white marble from this box is , making the probability of choosing Box 2, then a white marble, .

The overall probability of choosing a white marble is the sum of these probabilities:

Since both boxes contain the same number of marbles, the one with the most red marbles offers us the greater probability of drawing a red marble. For simplicity's sake, assume that each box contains thirty marbles - these arguments will be valid whatever the number.

From Statement 1, it follows that two-thirds of sixty, or forty, marbles total are red, but either box can contain more than the other.

From Statement 2, it follows that fifteen marbles in Box 2 are red, but no information about Box 1 follows.

From both statements, however, it follows that there are fifteen red marbles in Box 2 and twenty-five red marbles in Box 1, so Box 1 has more red marbles and offers the greater chance of drawing a red marble.

If only Statement 1 is assumed, there is no way of knowing which box offers the highest probability of drawing a red marble, since there is no way to tell which box has the higher portion of its marbles red without knowing how their numbers of white marbles or their total numbers of marbles compare. A smiliar argument holds for Statement 2 alone.

If both statements are known, however, each box has the same number of red marbles, and Box 2 contains more marbles overall - this makes Box 1 the box with the highest portion of its marbles red, and the box that offers the greater chance of drawing a red marble in a random draw.

A 3 can be rolled only by tossing a 1-2, but the loaded die can be either the 1 or the 2; similarly, an 11 can be rolled only by tossing a 5-6, but the loaded die can be either the 5 or the 6. We know only the probability of the loaded die coming up 2 and that of it coming up 5; we do not have this information for 1 or 6. We explore two cases and compare the probabilities of rolling a 3 and rolling an 11.

Case 1: Suppose the loaded die will come up 1 with probability 0.1 and 6 with probability 0.3. The probabilities of rolling a 3 or a 11 are, respectively:

and

making 11 the more likely outcome.

Case 2: Suppose the loaded die will come up 1 with probability 0.3 and 6 with probability 0.1. The probabilities of rolling a 3 or a 11 are, respectively:

and 

making 3 the more likely outcome.

Therefore, the two statements together provide insufficient information to answer the question.

Let  be the balls that are completely red, completely yellow, and bicolored, respectively; let  be the total number of balls. Then the probabilities of drawing a ball with some red and drawing a ball with some yellow are, respectively,

 and 

For the former probability to be greater than the latter, we can use some algebra to reveal:

That is, the former probability is greater than the latter if and only if there are more completely red balls than yellow balls. Similarly, the former probability is less than the latter if and only if there are fewer completely red balls than yellow balls. 

Therefore, Statement 1 is irrelevant; Statement 2, which compares the number of completely red balls and completely yellow balls, gives us our necessary and sufficient information.

The difference of the two dice will be 1 in case any of the following outcomes occur:

This is ten outcomes out of a possible 36, so this yields a probability of 

.

Remember, you do not have to SOLVE the data sufficiency problems.  In other words, you do not have to calculate the exact probability of getting a kirsch-flavored (cherry liquor) treat.  You just need to determine how many of the statements are NECESSARY to derive the answer.

The first statement by itself is not sufficient, because knowing that half of the treats are peanut butter flavored does not tell you the proportion that are kirsch flavored.

The second statement is not sufficient, because comparing the proportion of kirsch treats to peanut-butter treats won't help you find the probability of getting a kirsch treat unless you know the ratio of peanut-butter treats in the case.

But, if we put the two pieces of information together, we can solve the problem.  Statement 1 tells us the proportion of peanut-butter treats, and statement 2 tells us how many kirsch treats there are compared to peanut-butter treats.

Therefore, the correct answer is C.

Let P(A) be the probability of Anna picking an apple and P(O) be the probability of Anna picking an orange. We need to evaluate each statement individually first:

(1) gives us the total number of fruits, but no indication as to the number of each fruit, so we cannot determine probabilities form (1) alone

(2) gives us the proportion of oranges. Let N be the total number of fruits, A the number of apples and O the number of oranges. We have:

 We also know that .

We can rewrite this last equation as follow:

 or 

We know that P(A) = A / N and from statement (2) we get

 which we can simplify to find P(A)

Therefore the right answer is B.