Probablity of Being Chosen There are nine employess. Five are male, four are female.  Three employees will be selected to go to a conference.


*

*How many possible combinations of three employees are possible?

*If each employee has equal chance of being selected for the trip, what is the probability that all three selected are males?

 A: *

*Assuming order doesn't matter:

$\left( \begin{array}{c} 9 \\ 3 \end{array} \right) = \frac{9!}{6!\cdot3!}=\frac{9\cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} = 3\cdot 4 \cdot 7 = 84$


*The probability that we get a male upon randomly selecting an employee: $\frac{5}{9}$. If we were to select another male, however, we'd have a $\frac{4}{8}$ chance. For a third male, it'd be $\frac{3}{7}$.
Multiply the three probabilities together.
A: Hint 

*

*I assume we can tell apart our employees (would be a shame if not).

In how many ways can we chose the first employee?, the second?, the third?
  
  Since we can tell them apart, the choices are (almost) independent. We can multiply the results to get the full number. Note that we may not chose an employee twice, wich is why they are only "almost" independent. (That is, the number is independent of the prior choices, not the available individuals). Finally note that the order is irrelevant so we must divide by $3!$.



*Now we have fewer to select from, but the same process


In how many ways can we chose those three men?
  
  The probability is then given by the answer divided by the answer to 1.

