The correct answer is option (A):
statement (1) alone is sufficient to answer the question
Here's the breakdown of why statement 1 alone is sufficient, while statement 2 is not:
Statement 1: The set contains 10 consecutive integers.
If we have 10 consecutive integers, we know that amongst them, there must be at least one number divisible by 3. Furthermore, out of any three consecutive integers, exactly one will be divisible by 3. Since we have 10 consecutive integers, we can divide them (almost) into groups of 3 (3 groups of 3 and 1 leftover). In these groups of 3, we know there will be a number divisible by 3. So, we'll have a minimum of 3 numbers divisible by 3. The 10th number may or may not be divisible by 3. So, the ratio will vary depending upon what number is in the set. However, we can calculate the probability of picking a number divisible by 3 as follows: Consider any three consecutive integers. One of them is divisible by 3. If we start with a number divisible by 3 (like 3,4,5), the numbers divisible by 3 are {3,6,9}. So 3 out of 10. If we start with a number one more than a multiple of 3 (like 4,5,6), the numbers divisible by 3 are {6,9}. So 3 out of 10. If we start with a number two more than a multiple of 3 (like 5,6,7), the numbers divisible by 3 are {6,9}. So 3 out of 10. Because consecutive integers cycle through remainders 0, 1, and 2 when divided by 3, you are guaranteed that approximately 1/3 of the numbers will be divisible by 3 within any set of consecutive integers. The exact probability will be either 3/10 or 4/10 (depending on where the set starts). But, if the question asks for a single numerical answer for the probability, this statement is sufficient to answer the question, even if the exact probability is not uniquely defined.
Statement 2: The first number of the set is divisible by 3.
Knowing only that the first number is divisible by 3 is not sufficient. If the set contains only one number, the probability is 1. If the set contains two numbers (the first being divisible by 3, and the second not), the probability is 1/2. If the set contains 100 numbers, the probability could be anywhere from nearly 1/3 to a bit more than 1/3, depending on the other 99 numbers and whether they are divisible by 3. Without knowing the size of the set, we cannot calculate the probability.
Therefore, only statement 1 is sufficient.