[Guest]

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?

[bonanova]

.6

[CaptainEd]

6/9 = .67

edit: but I missed a case. I'm with Bonanova now...

Edited March 25, 2009 by CaptainEd

[HoustonHokie]

There are three different ways the workspaces can be assigned: 1, 2, or 3 men in offices. Because we're not talking about specific individuals, I think that each would be equally likely, and the case we're dealing with would have a probability of 1/3. But, I'm probably wrong (at least 50% chance of that

).

And, thinking about that assertion for a moment before I hit the Add Reply button, I think I probably am wrong. If the men are M1, M2, and M3 and the women are W1 and W2, then I think you can come up with 10 different ways to arrange them in workspaces:

Office: M1 M2 M3 Cube: W1 W2

Office: M1 M2 W1 Cube: M3 W2

Office: M1 M2 W2 Cube: M3 W1

Office: M1 M3 W1 Cube: M2 W2

Office: M1 M3 W2 Cube: M2 W1

Office: M2 M3 W1 Cube: M1 W2

Office: M2 M3 W2 Cube: M1 W1

Office: M1 W1 W2 Cube: M2 M3

Office: M2 W1 W2 Cube: M1 M3

Office: M3 W1 W2 Cube: M1 M2

6 of these arrangements have 2 men and 1 woman in offices, so the probability is 60%. There, I think that's more right

[bonanova]

First 3 are offices.

111 - 00

110 - 10

110 - 01

101 - 10

101 - 01

100 - 11

011 - 10

011 - 01

010 - 11

001 - 11

[Guest]

P(MMW) = 3/5 * 2/4 * 2/3 = 12/50

P(MWM) = 3/5 * 2/4 * 2/3 = 12/50

P (WMM) = 2/5 * 3/4 * 2/3 = 12/50

P(2 men/1 woman) = 36/50 = 0.72

Something tells me my reasoning may be flawed.

[Guest]

.6

Looks like Bonanova was first to get the correct answer. I swiped this from the "SAT question of the day", their explanation is below.

Of the 5 office workers, 3 are to be assigned an office. This is an

example of combinations: to find the number of ways of choosing 3 of

the 5 workers, you can count the number of ways of selecting the

workers one at a time and then divide by the number of times each

group of 3 workers will be repeated.

There are 5 ways of choosing the first worker to get an office. Then

there will be 4 ways of choosing the second worker to get an office,

and 3 ways of choosing the third worker. This is a total of 5 × 4 × 3

= 60 possibilities. In these 60 possible selections, each distinct

group of 3 workers will occur 3 × 2 × 1 = 6 times. (There are 3

possibilities for the first worker chosen from the group, 2 for the

second worker chosen, and only 1 for the third.) Therefore, there are

60 over 6 = 10 different ways the 3 workers who get an office can be

chosen from the 5 workers.

How many of these 10 possible groups of 3 workers consist of 2 men and

1 woman? From the 3 male workers, 2 can be chosen in 3 different ways.

There are 2 possibilities for the female worker. Therefore, 3 × 2 = 6

of the groups of 3 workers consist of 2 men and 1 woman.

Since there are 10 different ways the 3 workers who get an office can

be chosen, and 6 of these possible groups of 3 workers consist of 2

men and 1 woman, the probability that the offices will be assigned to

2 men and 1 woman is 6 over 10, or 3 over 5.

I got the same answer by enumerating the 20 combinations of people ABCDE in the two cubicles.

[Guest]

⁵C₃ - Total ways of choosing 3 ppl out of 5 for the office; ³C₂ - No. of ways of choosing 2 men out of 3 and ²C₁ - No. of ways of choosing 1 woman out of 2. So the probability is (³C₂*²C₁)/⁵C₃ = 3/5 = 0.6

Edited March 25, 2009 by johnogdenjunk

[Guest]

P(MMW) = 3/5 * 2/4 * 2/3 = 12/50

P(MWM) = 3/5 * 2/4 * 2/3 = 12/50

P (WMM) = 2/5 * 3/4 * 2/3 = 12/50

P(2 men/1 woman) = 36/50 = 0.72

Something tells me my reasoning may be flawed.

Miscalculated;

It's actually 12/60 * 3 = 36/60 = 0.6

My bad.

[Guest]

Looks like Bonanova was first to get the correct answer. I swiped this from the "SAT question of the day", their explanation is below.

I got the same answer by enumerating the 20 combinations of people ABCDE in the two cubicles.

assign the cubicles first. Only two possibilities exist for assigning the first cubicle M(3/5) or W(2/5). For each of these, only one second assignment will provide the necessary distribution: M(3/5)W(2/4) or(+) W(2/5)M(3/4) = 3/5.