tag:blogger.com,1999:blog-7691251033406320222.post1795206912272673410..comments2024-03-26T04:19:38.862-07:00Comments on kitchen table math, the sequel: SAT questionCatherine Johnsonhttp://www.blogger.com/profile/03347093496361370174noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-7691251033406320222.post-39271960805288183892010-09-09T18:11:28.529-07:002010-09-09T18:11:28.529-07:00How about:
Offices 1, 2, and 3 can be assigned to
...How about:<br />Offices 1, 2, and 3 can be assigned to<br />M-1, M-2, W-3, so 3/5 x 2/4 x 2/3 = 1/5<br />M-1, W-2, M-3, so 3/5 x 2/4 x 2/3 = 1/5<br />W-1, M-2, M-3, so 2/5 x 3/4 x 2/3 = 1/5<br />Add them and you get 3/5.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-63340202718534904852010-08-23T19:03:29.502-07:002010-08-23T19:03:29.502-07:00OM,
Your sample space is wrong.
You need to labe...OM,<br />Your sample space is wrong. <br /><br />You need to label all of the men, and all of the women, and then discount the ordering later, but you can't discount the labels in the first place.<br /><br />Solving it even with your precondition that one is a man, you're left with: how many total ways are there to put 2 people in the remaining 2 slots, and of them, how many have 1 man and 1 woman?<br /><br />Writing down all of the Ms and Ws you don't get what you described. Just considering M1 as the one you grab (and then recognizing order is irrelevant later):<br />you've got<br />M1 M2 M3, M1 M2 W1, M1 M2 W2, M1 M3 W1, M1 M3 W2, M1 W1 W2. Next, considering M2 (but not counting redundant ones) is M2 M3 W1, M2 M3 W2, M2 W1 W2. And last, M3 W1 W2.<br /><br />6 of 10 of those fit our defn. <br /><br />Your occupancy of office/cubicle doesn't consider all of those, because it only counts two times getting a man and a woman, but if you label them all out there are 4. so it's not 2 out of 4, it's 4 out of 6.<br /><br />It's as if you jumped into thinking that you were flipping coins to assign people to rooms, with each person getting a heads-office or tails-cubicle. But our events aren't independent like that, because of our occupancy numbers. So your model is wrong.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-18290810598748864802010-08-23T13:47:16.879-07:002010-08-23T13:47:16.879-07:00So, there are 5 people total, and 3 spaces in the ...So, there are 5 people total, and 3 spaces in the office. Since it doesn't matter which man or woman gets in, there are 5 possible "draws," out of which 3 include one woman and two men. Or something...Weston CT SAT prephttp://www.satprepct.com/college-tests/weston-ct-sat-prep.phpnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-30141810535638546052010-08-23T10:02:59.585-07:002010-08-23T10:02:59.585-07:00My great logic for the wrong answer:
a. 1 male m...My great logic for the wrong answer:<br /><br />a. 1 male must be in the office => prob = 1<br />b. for the other two males or two females, the sample space = oo, oc, oc, cc => prob = 1/2 of oc<br />c. Overall probability = 1 * 1/2 * 1/2 = 1/4.<br /><br />OK, where am I wrong?orangemathhttps://www.blogger.com/profile/05099727076265177042noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-45144638090237617612010-08-22T23:38:50.090-07:002010-08-22T23:38:50.090-07:00Oh, and yes, your friend's son's approach ...Oh, and yes, your friend's son's approach is simply wrong.<br /><br />If this was a real SAT question, it was a poor one. They should have chosen numbers so you couldn't bumble into the right answer with impossibly wrong thinking.<br /><br />So see how it's wrong, change the numbers:<br /><br />Of 7 employees, three are to be assigned an office and 4 are to be assigned a cubicle. If 5 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?<br /><br />2/7+1/7 = 3/7. But that's not the probability for the event stated here, which is 4/7. (try using the method of counting how many ways there are to choose 3 people from 7, then 2 of those 3 being men, and 1 being a woman.)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-5181512476343461692010-08-22T22:44:29.117-07:002010-08-22T22:44:29.117-07:00This is called a hypergeometric distribution. Supp...This is called a hypergeometric distribution. Suppose that an urn contains n marbles of which m are white, and we select r marbles from the urn. Then the probability that k of the marbles in the sample are white is <br /><br />C(m,k)*C(n-m,r-k) / C(n,r).<br /><br />We get the answer 3/5 by substituting 5,3,3,2 for n,m,r,k.<br /><br />I don't know if this type of problem appears frequently enough on the SAT to justify learning this formula, but I thought that I would mention it anyhow.Davidhttps://www.blogger.com/profile/09232747857608296294noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-59721507768111590642010-08-22T21:27:20.534-07:002010-08-22T21:27:20.534-07:00Did you try to do it with n choose k?
What's ...Did you try to do it with n choose k?<br /><br />What's the total number of ways for putting 3 people into the office? <br /><br />It's 5 choose 3: of the 5 people, choose 3. <br /><br />Then, the probability of picking the event specified by the problem (2 men, 1 woman) is the number of ways of that event occurring out of the total number of ways of putting 3 in the office.<br /><br />what's the total number of ways of putting 2 men and 1 woman in the office?<br /><br />3 choose 2 * 2 choose 1<br /><br />do you know why can you multiply those two parts?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-75249321184618911462010-08-22T18:27:13.194-07:002010-08-22T18:27:13.194-07:00Think about the women being placed sequentially an...Think about the women being placed sequentially and multiply probabilities.<br /><br />For exactly one women to get an office, either the 1st women gets an office and the 2nd woman does not, with probability <br /><br />3/5 * 2/4<br /><br />or the 1st woman gets a cubicle and the 2nd gets an office, with probability<br /><br />2/5 * 3/4 .<br /><br />(It is not a coincidence that these probabilities are the same). Adding 6/20 to 6/20, one gets 3/5.Bostoniannoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-40414511402404914552010-08-22T17:31:34.494-07:002010-08-22T17:31:34.494-07:00Here's my question.
My friend's son solve...Here's my question.<br /><br />My friend's son solved this problem by adding 2/5 (2 men out of 5 people) to 1/5 (1 woman out of 5 people).<br /><br />That produces the right answer but it seems like the wrong approach --- <br /><br />Although I have carefully worked my way through Dolciani's chapter on probability, I couldn't do this problem.<br /><br />Sad to say.<br /><br />I was close.<br /><br />I need to go back and re-do all the problem sets. I've only done them once.Catherine Johnsonhttps://www.blogger.com/profile/03347093496361370174noreply@blogger.com