If a 3-digit integer is selected at random from the integers 700 through 899, inclusive, what is the probability that the second digit is one greater then the first digit, and the third digit is one greater than the second digit?
Options:
(A) 2/225
(B) 2/200
(C) 8/700
(D) 1/50
(E) 1/200 Hmmm, my stream of consciousness:
Hundreds digit can be 7 or 8
If the hundreds digit is 7, then the tens digit has to be 8 and the units digit has to be 9
If the hundreds digit is 8, then the tens digit has to be 9 and the units digit has to be 10
A digit has to fall between 0 and 9, so 10 cannot be a digit
The number is 789
Total numbers between 700 and 899 = (899-700)+1 = 200
Probability = 1/200 So, the answer is (E).