<p>@dkdkdk</p>
<p>yes but the question didnt specify that it mattered which person got what office, as long as they are the same gender.</p>
<p>i dunno I am really bad at this kind of probability too…lucky i didnt get too many last test, i got a 800.</p>
<p>^Guys, if you are having trouble visualising this stuff, write out: __ __ __. These represent the three offices (fill each of these with the corresponding probabilities). Now, the probability of the first office being filled by a man is 3/5. Then, the next office has to be filled with either a man or a woman. The probability that the second office is filled by a man is 2/4 (subtracted 1 man from numerator and 1 person from denominator). The third office must be filled by a woman, which is a 2/3 probability. All three occuring is a (3/5)(2/4)(2/3)=4/20=1/5. There is also the occurence that a man fills the first office but a woman fills the second office and a man fills the third. All three of these occuring is (3/5)(2/4)(2/3)=4/20=1/5. There is also the chance that a woman fills the first office, a 2/5 probability, and then a man must fill the second office (3/4 prob.) and another man must fill the third (2/3 probability). All three occuring is a (2/5)(3/4)(2/3)=4/20=1/5. Since all three methods of filling the offices are possible, you add these probabilities to get 1/5+1/5+1/5=3/5. </p>
<p>^Granted, this takes somewhat longer than the choosing method some of you haven’t learned yet. But, you can simplify the calculations by reasoning that the probability of 2 men and 1 woman being assigned to offices is the same probability as 1 man and 1 woman being assigned to cubicles. </p>
<p>These problems aren’t impossible, you just have to boil them down to their corresponding parts. In my opinion, actually having blanks that you can fill in with probabilities (as if you were assigning the offices in real life) makes the problem easier to understand.</p>
<p>Gamma that is a great explanation, and should be what you do if you haven’t learned the other method</p>