<p>This question was an SAT question of the day about 2 weeks ago:</p>
<p>Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. 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</p>
<p>b) 2/5</p>
<p>c) 1/2</p>
<p>d) 3/5</p>
<p>e) 2/3</p>
<p>Here's what i did: The probability that it'll be 2 men is 2/3 (because there are 3 men and 2 of those three can be assigned to the office). The probability that it'll be 1 woman is 1/2 (out of the 2 women one could be assigned to the office). I multiplied both probabilities since the question states "probability that it'll be 2 men AND 1 woman," and get A: 3/6 simplified is 1/2.</p>
<p>Consider how many ways the men and women can be arranged to be equivalent to a five letter word consisting of three M's and two W's.</p>
<p>The number of five letter words is 5!</p>
<p>Since the three M's are indistinguishable and the two W's are indistinguishable, the 5! needs to be divided by 3! x 2!.</p>
<p>5! = 120
3! = 6
2! = 2</p>
<p>so there are 10 (=120/12) unique 5 letter words consisting of only three M's and two W's (You could also write the list of ten words pretty quickly).</p>
<p>We want to know how many of these ten words are "good" -- namely, how many have two M's and a W together. (Again, if we had written out the list, we could count good or bad directly).</p>
<p>For convenience, let the first three spaces be the office and the last two spaces be the cubicle.</p>
<p>By inspection, we see that having two women in the cubicle (MMM-WW) or two men in the cubicle (_ _ _ - MM) won't work. There is exactly one word of the form MMM-WW and are exactly three words of the form _ _ _ - MM, so four words don't work.</p>
<p>Four bad, ten total, so six good;
6/10 reduces to 3/5</p>
<p>hey guys, to solve this question, i actuallly used two different approaches. It's actually quite easy once you get the question and with more practice of course. :)</p>
<p>Now the question says "Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. 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>Approach 1:
When I looked at this problem, i asked myself, how many people are there? Well, the answer is obviously 5. Now, the question wants 2 men and 1 women in the office. So the chance of getting the first man into the office is 3/5 (because there are simply 3 men out of your sample of 5 people!). </p>
<p>The chance of getting the second man into the office is 2/4. You may ask why 2/4? Well, simply, because since one man has been used up, there are 2 men left out of the remaining 4 people!</p>
<p>So now we have 2 men in the office. Now for the women. At this point, how many people are there left to choose from? The answer is 3 of course! How many women are there out of these 3 people left? 2! So the chance of getting the women into the final office slot is 2/3!</p>
<p>Ok before you actually multiply 3/5 X 2/4 X 2/3 , take note that this is NOT the answer! you may ask why? Well simply, the arrangement of the 2 men and 1 woman in the office is MMW. Remember that you can arrange the MMW into other arrangements, WMM, MWM. So there are 3 (3!/2! --> ask you teacher if you don't understand why i wrote this) arrangements altogether. </p>
<p>Therefore, the answer is simply 3/5 X 2/4 X 2/3 X THREE = 3/5!</p>
<p>Hmm don't ask why i did this, since it's rather difficult to explain. The first approach is the recommended approach though. If you guys have other difficult math problems, PM me, i'll be more than willing to help! :)</p>
<p>spideyunlimited has provided a good approach as well, using the combinatorics method. It is up to you which one to use. But for beginners try using the probability approach since it can work for many cases.</p>
<p>thanks kwaldner
your explanation helped me
i haven't yet been in algebra II so i don't understand the "3!/2!"
but i'm sure i will next year
thanks again</p>
<p>OK - These answers are fine, but always remember that SAT tests logic and reasoning as much as it does content knowledge. So there's usually a logical and not too complicated way to figure these things out, even if you don't know the formulas to use. Check this out:</p>
<p>If the offices break two men and one woman, then that means the cubicles have to have one man and one woman (use cubicles because two people are easier to figure than three).</p>
<p>List all the possible two-person cubicle combinations - the first person with each of the remaining four; the second person with people three through five; the third person with people four and five; and finally, numbers four and five together:</p>
<p>^^dillbilly123, gadad just posted a great way to solve the problem, by listing possibilities for the 2-person cubicle. A related way to solve the problem, again focusing on the cubicle, which involves just 2 people--so it's easier--is:
1) Assign the first person to the cubicle. The odds are 3/5 that it's a man, 2/5 that it's a woman.
2) Now assign the second person to the cubicle. Notice the that probabilities are different for the second assignment, depending on what the first assignment was. If the first was a man, you have 4 people left, 2 men and 2 women. So the odds are 2/4 = 1/2 that a woman is assigned. If the first person assigned was a woman, you again have 4 people left, but 3 are men and 1 is a woman, so the odds that the second person is a man is 3/4. The overall probability is
3/5 * 1/2 + 2/5 * 3/4 = 3/5.</p>
<p>^^dillbilly123, It would help you also to figure out what was wrong with your approach? Looking at what you did, the essence of your difficulty seems to be that you need to figure out exactly how to apply the counting arguments in probability. </p>
<p>You clearly have the basic idea that the probability is the number of ways that a particular outcome could be observed, divided by the total number of possible outcomes. The key to getting the probability questions right, though, is to understand what "particular outcome" and "total number of possible outcomes" should be involved in the calculation, and how.</p>
<p>You are trying to determine the probability that 2 men and 1 woman will be in the offices, given 3 men and 2 women to be assigned to offices or cubicles. So, you start by saying that the probability that it will be 2 men is 2/3, because there are 3 men. To see why this is not right, suppose that the question asked for the probability that 3 men would be in the offices, given 3 men and 2 women to be assigned to offices or cubicles. Do you see that the number of men should not be in the denominator? </p>
<p>The probability to assign a woman is not 1/2, in this case (unless 1 man has already been assigned to the office). If you had two women (Doris and Chloris), and you knew that one had been assigned an office randomly, the probability that it was Doris is 1/2. But that's not what this question is asking. So the number of women shouldn't be in the denominator, either.</p>
<p>Another way to look at it--for the purpose of understanding the probability calculation now, so that you can handle it on the SAT later, not for the purpose of going through this train of thought on the actual test--is to think about the case where there are 3 men and 200 women, and 3 people are going to be assigned to offices. The odds of 2 men being assigned to the offices would (pretty obviously) not be 2/3 in this case.</p>