Nice twist of similar questions. There are 7*6*5=210 possible 3-digit numbers formed by digits 1-7 without repetition. For numbers greater than 700, the first digit must be 7. Of the 30, the last digit must be a 2,4, or 6 to be an even number, that leaves 3 possibilities for the last digit. The middle digit can be any number of the remaining 5 digits. So the number of qualifying candidates would be 1*5*3=15 (1 for the first digit, 5 choices for the second, and 3 choices for the last digit). Probability of randomly choosing a qualifying candidate (>700 & even) is therefore 15/210=1/14
