SOLUTION: The finance department of a particular company has 15 employees, 6 of whom have MBA degrees. Suppose we select three different employees sequentially at random from this department

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Question 1146892: The finance department of a particular company has 15 employees, 6 of whom have MBA degrees. Suppose we select three different employees sequentially at random from this department. Determine the probability of the following events.
a) The first employee has an MBA, given that there is a total of one MBA among all three employees.
b) There are exactly two employees with an MBA, given that the first employee has an MBA.
c) The first employee has an MBA, given that there is at least one MBA among all three employees.
Found 3 solutions by womansecret13, ikleyn, Edwin McCravy:
Answer by womansecret13(1) About Me  (Show Source):
Answer by ikleyn(52786) About Me  (Show Source):
Answer by Edwin McCravy(20056) About Me  (Show Source):
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Here are the 8 cases and their enumerations. (let M indicate that an MBA is
selected and N indicate that a Non-MBA is selected.

                                    The      Is the first  Are there    At  
                                   exact       selected    exactly 2  least
                                no. of MBAs     an MBA?     MBA's?    1 MBA?
----------------------------------------------------------------------------
#1  n(M M M) = (6)(5)(4) = 120       3           yes         no        yes 
#2  n(M M N) = (6)(5)(9) = 270       2           yes        yes        yes
#3  n(M N M) = (6)(9)(5) = 270       2           yes        yes        yes
#4  n(M N N) = (6)(9)(8) = 432       1           yes         no        yes
#5  n(N M M) = (9)(6)(5) = 270       2            no        yes        yes
#6  n(N M N) = (9)(6)(8) = 432       1            no         no        yes
#7  n(N N M) = (9)(8)(6) = 432       1            no         no        yes
#8  n(N N N) = (9)(8)(7) = 504       0            no         no         no
----------------------------------------------------------------------------
        15P3 = 15∙14∙13 = 2730   <--added as a check
a) The first employee has an MBA, given that there is a total of one MBA
among all three employees.
n(#4)/n(#4 or #6 or #7) = 432/(432+432+432) = 1/3
b) There are exactly two employees with an MBA, given that the first
employee has an MBA.
n(#2 or #3)/n(#1 or #2 or #3 or #4) = (270+270)/(120+270+270+432) = 540/1092
= 45/91
c) The first employee has an MBA, given that there is at least one MBA among
all three employees.
n(#1 or #2 or #3 or #4)/n(#1 or #2 or #3 or #4 or #5 or #6 or #7) = 
(120+270+270+432)/(2730-504) = 1092/2226 = 26/53

Edwin