<p>Here it is, thanks for your help!</p>
<p>Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a crucible. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?</p>
<p>A. 1/3
B. 2/5
C. 1/2
D. 3/5
E. 2/3</p>
<p>Try finding the probability that the offices will not be assigned to 2 men and 1 women - then subtract that from 1 to get the answer.</p>
<p><a href=“http://talk.collegeconfidential.com/sat-preparation/530894-math-question.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/530894-math-question.html</a></p>
<p>If 2 men and 1 woman are in the offices, that means one man and one woman need to go into the cubicles. So, for the first cubicle, there is a 3/5 probability of filling it with a man. For the second cubicle, there is a 2/4 probability of filling it with a women (denominator changes because one man is already in a cubicle, thus only 4 people remain to be assigned). The probability of both is (3/5)<em>(2/4)=6/20=3/10. OR, you could fill the first cubicle with a woman, which is a 2/5 probability and the second with a man, a 3/4 probability. The probability of that is (2/5)</em>(3/4)=6/20=3/10. Then, since both combinations are possible, you add those probabilities together to get 6/10=3/5. This is the probability that 1 man and 1 woman get in the cubicles, which also forces 2 men and 1 woman to be in the offices. So the probability is D, 3/5.</p>
<p>^I think I’m right. This is kind of a weirdly worded problem.</p>