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.
Thanks in advance
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.
Thank you all. I really appreciate the assistance!
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.)