Question 1146892
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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?
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#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
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        15P3 = 15∙14∙13 = 2730   <--added as a check</pre>a) The first employee has an MBA, given that there is a total of one MBA
among all three employees.<pre>n(#4)/n(#4 or #6 or #7) = 432/(432+432+432) = 1/3</pre>b) There are exactly two employees with an MBA, given that the first
employee has an MBA.<pre>n(#2 or #3)/n(#1 or #2 or #3 or #4) = (270+270)/(120+270+270+432) = 540/1092
= 45/91</pre>c) The first employee has an MBA, given that there is at least one MBA among
all three employees.<pre>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</pre>