Final answer:
The probability of forming a three-digit, even number greater than 700 using the digits 1 to 7 without repetition is 1/14.
Step-by-step explanation:
The question revolves around finding the probability of forming a three-digit number using the digits 1, 2, 3, 4, 5, 6, and 7, with the conditions that the number must be even and greater than 700. First, to be even, the number must end in an even digit (2, 4, or 6). Second, to be greater than 700, the first digit must be 7. Since we cannot repeat digits, we are left with five choices for the middle digit after choosing 7 for the first digit and an even digit for the last. Therefore, the total number of favorable outcomes is 3 (options for the last digit) multiplied by 5 (options for the middle digit).
To determine the total number of possible three-digit numbers we can form from the seven available digits without restrictions apart from no repetition, we observe that we have 7 choices for the first digit, 6 for the second digit, and 5 for the third digit, leading to 7 x 6 x 5 total outcomes. The probability of picking a number that is both even and greater than 700 is thus the number of favorable outcomes divided by the total number of outcomes. This yields a probability of (3 x 5)/(7 x 6 x 5), which simplifies to 3/42, or 1/14.