A division of a company consists of seven men and five women. If two of these twelve employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?	Options:
A) 1/6
B) 2/5
C) 2/9
D) 5/12
E) 5/33
First, the denominator. We have twelve different people, and we want a combination of two selected from these twelve. We will use the formula:   nC2 = n(n-1)/2   which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n – 1) positive integers. Here   12C2=12(11)/2=6*11= 
<gadget id="calculator">6*11</gadget>
<output>66</output>
66   That’s the total number of pairs we could pick from the twelve employees. That’s our denominator.   For the numerator, we want every combination of two from the five female employees. That’s   5C2=5(4)/2=5*2= 
<gadget id="calculator">5*2</gadget>
<output>10</output>
10   That’s the number of pairs of female employees we could pick from the five. That’s our numerator.   probability= 10/66 =5/55   Answer = 5/33
<result>E</result>
A division of a company consists of seven men and five women. If two of these twelve employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?