<p>Can anyone explain to me how to do this problem? Thanks.</p>
<p>Of 5 office workers, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the workers 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>I would also appreciate an explanation. Anyone!</p>
<p>Is the answer 1/10?</p>
<p>it should be 3/5 i think. i mean if there are 5 ppl, 2/5 men + 1/5 women for a total of 3/5. I think they word it quite deceptivley</p>
<p>do you have the choices??</p>
<p>I think the answer is 3/5, but that's not the way to do it. I think it's a combination and a probability question. I'll post the solution it later when I have time and if no one else does before me.</p>
<p>1/2.</p>
<p>Look only at the women.</p>
<p>If the women are assigned randomly,</p>
<p>the probability of two women to the office is (.5)(.5) = .25.</p>
<p>the probability of two women to the cubicle is (.5)(.5) = .25.</p>
<p>the remaining probability, (1 - .25 -.25 = .5), is the probability of one woman to the cubicle and one woman to the office.</p>
<p>Yeah, I think it is 6/10 or 3/5. I did 3 C 2 * 2 C 1= 6/10</p>
<p>It's like so: You have to choose 3 workers out of a total of 5 workers, so 5C3, which is 10 and you must choose 2 men out of 3 men, and 1 woman out of 2 women, which becomes: 2C1*3C2, which is 6. And 6/10 is 3/5.</p>
<p>Thanks for helping!</p>
<p>it is 3/5, i remember this from the Question of the day</p>