What is the probability that an integer \( (1 \le x \le 25) \) chosen at random is divisible by both 2 and 3?

To find the probability that an integer between 1 and 25 is divisible by both 2 and 3, we need to count the number of integers that satisfy this condition, and divide by the total number of integers between 1 and 25. A number is divisible by both 2 and 3 if and only if it is divisible by their product, which is 6. Therefore, we need to count the number of integers between 1 and 25 that are divisible by 6. To do this, we can start by finding the smallest multiple of 6 that is greater than or equal to 1, which is 6 itself. The next multiples of 6 are 12, 18, and 24, all of which are less than or equal to 25. Therefore, there are 4 integers between 1 and 25 that are divisible by both 2 and 3. The total number of integers between 1 and 25 is 25, so the probability that an integer chosen at random from this range is divisible by both 2 and 3 is 4/25. Therefore, among the given options, the answer is the one represented by the fraction 4/25.