jack and jill work at a hospital with 4 other workers. for an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. what is the probability that jack and jill will both be chosen?	Choices:
A) 1 / 3 B) 1 / 4 C) 1 / 15 D) 3 / 8 E) 2 / 3
<gadget id="calculator">factorial(6)</gadget>
<output>720</output>

<gadget id="calculator">factorial(2)</gadget>
<output>2</output>

<gadget id="calculator">factorial(4)</gadget>
<output>24</output>

<gadget id="calculator">2 * 24</gadget>
<output>48</output>

<gadget id="calculator">720 / 48</gadget>
<output>15</output>

<gadget id="calculator">1 / 15</gadget>
<output>1/15 = around 0.066667</output>

<result>C</result>
jack and jill work at a hospital with 4 other workers. for an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. what is the probability that jack and jill will both be chosen?
inverse(divide(factorial(6), multiply(factorial(2), factorial(4))))
"total number of ways to choose 2 out of 6 workers = 6 ! / 2 ! 4 ! = 15 number of ways to choose both jack and jill = 1 probability = 1 / 15 c should be the answer"