Would someone please show me how to find P(M1), P(F2), and P(M3) from the problem below
Problem: A company employs six male executives and three female executives. Three executives are randomly selected, one at a time, from the nine executives; the first one selected is sent to China, the second one is sent to New York, and third one is sent to France. Using the general multiplication rule, find the probability that a male is sent to China, a female is sent to New York, and a male is sent to France. Express your answer as a reduced fraction. Hint: Write the probability as P(M1F2M3), then apply the rule.
Here is how I approach the solution:
1. the Sample space S = 9(8)(7) = 504 ways to choose the 3 executives randomly
2. P(M1F2M3) = P(M1)*p(F2|M1)*P(M3|M1F2)
3. But I am not sure how to find P(M1) or P(F2) or P(M3)
Thank you very much for the help Plato.
Yes, I think I can finish the rest of the problem.
I hope my P(M3) is correct
Here is my answer
P(M3) = 5/7
P(M1F2M3) = P(M1)P(F2|M1)P(M3|M1F2) = 2/3*3/8*5/7 = 30/168 = 5/28
P.S. I really like your math display. You must use Math Type Program or something. I have Math Type but I could not copy and paste on this page. Any idea how I can display my math symbol here? Thanks again.