If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the second digit is two greater than the first digit and the third digit is less than the second digit?  Choose the correct choice from the following.
A) 1/100	B) 3/100	C) 2/103	D) 2/150	E) 1/50
Hundreds digit has to be 1 . So tens digit needs to be 2 greater than 1   1+2= 
<gadget id="calculator">1+2</gadget>
<output>3</output>
3   The tens digit is 3, and the units digit must be less than the tens digit so the units digit must be less than 3   0, 1, and 2 are all less than 3   140, 141, and 142 are all acceptable numbers giving us 3 possible numbers   Total numbers between 100 and 199= (199-100)+1= 
<gadget id="calculator">(199-100)+1</gadget>
<output>100</output>
100   Probability= 3/100   Answer 3/100
<result>B</result>
If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the second digit is two greater than the first digit and the third digit is less than the second digit?