Probability question

de3as said:

From nine people, there is 5 men and 4 women. It will be chosen from among them a chief, a secretary and a treasurer. What is the probability of being chosen a male chief or a female treasurer?
a.13/16
b.13/17
c.13/18
d.13/19
e.13/20

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de3as said:

From nine people, there is 5 men and 4 women. It will be chosen from among them a chief, a secretary and a treasurer. What is the probability of being chosen a male chief or a female treasurer?

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If you think of the positions as being an ordered seating, you can think in terms of permutations.

You have nine people. How many can you choose for the first (that is, the "chief") position (that is, the chair on the left)? How many are left for the second (that is, the "secretary") position (that is, the middle chair)? How many are left for the third (that is, the "treasurer") position (that is, the chair on the right)?

So in how many ways can you choose the three people? This is the total number of possible choices of committee members.

Now consider the same situation, but restricting the choice for the first (left-most) chair to a man. In how many ways can you fill the three positions?

Now consider the same situation, but restricting the choice for the third (right-most) chair to a woman. In how many ways can you fill the three positions?

Note, however, that each of these restricted choices "contains" the other within it. If the chief is a man, the treasurer might be a woman. If the treasurer is a woman, the chief might be a man. Since you don't want to double-count:

Consider the situation in which the chief is a man and the treasurer is a woman. In how many ways can you fill the three positions?

Using these three restricted-choice arrangements, in how many ways can you choose the three people in qualifying ways?

Then what is the probability?

If you get stuck, please reply showing your work for each of the questions above. Thank you!