# Thread: Either/And probability

1. ## Either/And probability

Would someone please help me to solve this question. It's not an assigned problem, but I believe if I understand how it works, that I could do my homework on my own, which what I want more than anything.

Twelve people work together: four of them are managers and eight of them are office workers. Two people are chosen at random from the twelve members of the group. Calculate the probability that one is a manager and one is an office worker. Express your answer as a fraction.

2. Originally Posted by MadeInNY
Would someone please help me to solve this question. It's not an assigned problem, but I believe if I understand how it works, that I could do my homework on my own, which what I want more than anything.

Twelve people work together: four of them are managers and eight of them are office workers. Two people are chosen at random from the twelve members of the group. Calculate the probability that one is a manager and one is an office worker. Express your answer as a fraction.

Reflect on the various parts of the unsimplified answer to understand where it comes from: $\frac{^4C_1 \cdot ^8C_1}{^{12}C_2}$.

Now express it in simplest fraction form.

3. Thank you for the quick reply. I'm not getting the correct answer yet, but at least with your help, I think that I'll be able to figure it out.

4. I'm still not getting the correct answer.

Here's what I did:

$(4*8/12) * 2 = 16/3$

Then I tried:

$(4*8/12) * (4*8/11) = 265/33$

Neither answer that I plugged in was correct. What am I missing?

5. $\frac{^4C_1 \cdot ^8C_1}{^{12}C_2}=\frac{(4)(8)}{(6)(11)}$.

6. Originally Posted by Plato
$\frac{^4C_1 \cdot ^8C_1}{^{12}C_2}=\frac{(4)(8)}{(6)(11)}$.
I know where the 11 came from, but not seeing where the 6 is from.

EDITED: Am I correct in figuring that the 6 is a result of there being 2 events, so 12/2 = 6?

(Sorry...I'm really trying hard; I want to understand, not have others do my assignments)

7. Originally Posted by MadeInNY
Would someone please help me to solve this question. It's not an assigned problem, but I believe if I understand how it works, that I could do my homework on my own, which what I want more than anything.

Twelve people work together: four of them are managers and eight of them are office workers. Two people are chosen at random from the twelve members of the group. Calculate the probability that one is a manager and one is an office worker. Express your answer as a fraction.

If you are having trouble with the solution given by Mr. F (which I suggest is well worth your trouble working to understand), here is another approach.

There are 4 managers out of 12 people, so the probability that the first person chosen is a manager is 4/12. Then there are 11 people left, so the probability that the second person chosen is an office worker is 8/11. So the probability that the first person chosen is a manager and the second person is an office worker is (4/12) (8/11).

Find the probability that the first chosen is an office worker and the second is a manager the same way. (The result will be the same as 1-manger 2-office worker.) Then add the two probabilities together; that's your answer.

8. Originally Posted by awkward

There are 4 managers out of 12 people, so the probability that the first person chosen is a manager is 4/12. Then there are 11 people left, so the probability that the second person chosen is an office worker is 8/11. So the probability that the first person chosen is a manager and the second person is an office worker is (4/12) (8/11).
This was my initial logic, but the answer was still wrong; I wasn't getting "16/33". Plato's breakdown actually lead the correct answer, but I completely agree that I need to understand Mr. Fantastic's notations. Unfortunately, it comes later in the book, so I have to jump ahead and work my way back.

Thank you all. I really appreciate the assistance!

9. Originally Posted by mr fantastic
Reflect on the various parts of the unsimplified answer to understand where it comes from: $\frac{^4C_1 \cdot ^8C_1}{^{12}C_2}$.

Now express it in simplest fraction form.
Originally Posted by MadeInNY
I'm still not getting the correct answer.

Here's what I did:

$(4*8/12) * 2 = 16/3$

Then I tried:

$(4*8/12) * (4*8/11) = 265/33$

Neither answer that I plugged in was correct. What am I missing?
$^{12}C_2 = \frac{12!}{2! \cdot (12 - 2)!} = \frac{12!}{2! \cdot 10!} = \frac{(12)(11)}{2} = 66$.

10. Originally Posted by mr fantastic
$^{12}C_2 = \frac{12!}{2! \cdot (12 - 2)!} = \frac{12!}{2! \cdot 10!} = \frac{(12)(11)}{2} = 66$.
Thank you!!! I just got some extra help from my professor, so it's starting to make sense now.
My book did not explain this topic very well. I did not realize that it was supposed to be treated like two separate fractions.
I appreciate your help (all of you who responded.)