<p>this was a qotd a few months ago that i never really understood. i suck at combination/permutation problems.</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
B.2/5
C.1/2
D.3/5
E.2/3</p>
<p>ooh btw the answer is d)3/5</p>
<p>There are 3 office spaces. There are 3 possible ways we can select 2 men and a woman to occupy those offices. They are: 1) M-M-W, 2) M-W-M, 3) W-M-M. Lets write out the probabilities of each.</p>
<p>1) (3/5)<em>(2/4)</em>(2/3) = 12/60 (at the first selection, there are 3 men out of 5 people, so the probability is 3/5, there are 2 men left to “compete” for the second spot so the probability is 2/4, and then there are 2 women to be considered for the last spot, thus probability is 2/3)</p>
<p>2) (3/5)<em>(2/4)</em>(2/3) = 12/60 (by analogy)</p>
<p>3) (2/5)<em>(3/4)</em>(2/3) = 12/60 (by analogy)</p>
<p>The probability of selecting any 2 men and a woman for 3 office spaces is the sum of 1), 2), 3) and equals 3*(12/60) = 3/5.</p>
<p>Hope this helps.</p>
<p>You could also find the probabilty of having a man and a woman assigned to cubicles and then answer the original question by finding the complementary probability, i.e., by subtracting it from 1. Either way should give 3/5 as the answer.</p>
<p>I think the easiest way to do this is to list all the possibilities, then calculate probability from there.</p>
<p>What are the different combinations of people that can fill the 3 office spaces? Let the men be represented by A, B, and C; the women by 1 and 2.</p>
<p>3 person office possibilities
A B C
1 2 A
1 2 B
1 2 C
A B 1
A B 2
B C 1
B C 2
A C 1
A C 2</p>
<p>Notice that we have taken care of all possibilities (all men, two women and a man, two men and a woman). How many of those 10 possibilities consist of two men and a woman? The answer is 6.</p>
<p>6/10 = 3/5</p>
<p>you could also reverse it to make it simpler. You could say what is the probability that 1 man and 1 woman end up in the cubicle. The possibilities are:</p>
<p>A 1
B 1
C 1
A 2
B 2
C 2
A B
B C
C A
1 2</p>
<p>As you can see the top 6 represent 1 of each in a cubicle. So 6/10 = 3/5.</p>
<p>The problem with last 2 methods is that as the selection or population increases, situation will quickly get out of control, as it becomes very hard and tedious to make a list of combinations. For instance, imagine assigning 10 men and 3 women to 8 different offices such that there is 7 men and a woman selected to occupy offices. You will have to make a long list of combinations and then try to count them. However, for SAT, where the problems are short and simple, it is, of course, perfectly workable. ;)</p>
<p>That’s the point though. I doubt SAT problems would get that complicated, and writing out all the possibilities is great for reducing the chance of an error.</p>
<p>3c2<em>2c1 / 5c3
= 3</em>2 / (5*4/2)
= 3/5</p>
<p>3C2 for number of ways of choosing 2 men from the 3 present, 2C1 for choosing 1 woman from the 2 women present. And total possibilities of choosing any 3 persons randomly from the 5 persons is 5C3.
And of course, probability = no. of favorable cases / total no. of cases</p>