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Of 5 employees, three 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?
Answer Choices:
(A) 1 / 3
(B) 2 / 5
(C) 1 / 2
(D) 3 / 5
(E) 2 / 3
9 comments:
Here's my question.
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).
That produces the right answer but it seems like the wrong approach ---
Although I have carefully worked my way through Dolciani's chapter on probability, I couldn't do this problem.
Sad to say.
I was close.
I need to go back and re-do all the problem sets. I've only done them once.
Think about the women being placed sequentially and multiply probabilities.
For exactly one women to get an office, either the 1st women gets an office and the 2nd woman does not, with probability
3/5 * 2/4
or the 1st woman gets a cubicle and the 2nd gets an office, with probability
2/5 * 3/4 .
(It is not a coincidence that these probabilities are the same). Adding 6/20 to 6/20, one gets 3/5.
Did you try to do it with n choose k?
What's the total number of ways for putting 3 people into the office?
It's 5 choose 3: of the 5 people, choose 3.
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.
what's the total number of ways of putting 2 men and 1 woman in the office?
3 choose 2 * 2 choose 1
do you know why can you multiply those two parts?
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
C(m,k)*C(n-m,r-k) / C(n,r).
We get the answer 3/5 by substituting 5,3,3,2 for n,m,r,k.
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.
Oh, and yes, your friend's son's approach is simply wrong.
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.
So see how it's wrong, change the numbers:
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?
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.)
My great logic for the wrong answer:
a. 1 male must be in the office => prob = 1
b. for the other two males or two females, the sample space = oo, oc, oc, cc => prob = 1/2 of oc
c. Overall probability = 1 * 1/2 * 1/2 = 1/4.
OK, where am I wrong?
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...
OM,
Your sample space is wrong.
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.
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?
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):
you've got
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.
6 of 10 of those fit our defn.
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.
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.
How about:
Offices 1, 2, and 3 can be assigned to
M-1, M-2, W-3, so 3/5 x 2/4 x 2/3 = 1/5
M-1, W-2, M-3, so 3/5 x 2/4 x 2/3 = 1/5
W-1, M-2, M-3, so 2/5 x 3/4 x 2/3 = 1/5
Add them and you get 3/5.
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