If a 2-digit integer is selected at random from the integers 60 through 99, inclusive, what is the probability that the first digit is equal to two less than the second digit? Choose one
A) 4/40
B) 4/90
C) 60/100
D) 1/99
E) 4/60
The tens digit may be 6, 7, 8, or 9   The units digit may be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9   The tens digit has to be 2 less than the value of the units digit so we will add 2 to possible tens digits   6+2= 
<gadget id="calculator">6+2</gadget>
<output>8</output>
8   7+2= 
<gadget id="calculator">7+2</gadget>
<output>9</output>
9   8+2= 
<gadget id="calculator">8+2</gadget>
<output>10</output>
10   9+2= 
<gadget id="calculator">9+2</gadget>
<output>11</output>
11   Of these values, 8 and 9 are possible units digits which means that 6 and 7 are possible tens digits   Possible numbers are 68, 69, 78, and 79   We have 4 possible numbers   Total numbers between 60 and 99= (99-60)+1= 
<gadget id="calculator">(99-60)+1</gadget>
<output>40</output>
40   Probability= 4/40   4/40nswer 4/40
<result>A</result>
If a 2-digit integer is selected at random from the integers 60 through 99, inclusive, what is the probability that the first digit is equal to two less than the second digit?