nine candidates for hire , six men, three women,only two position list probability outcome?

Question

nine candidates for hire , six men, three women,only two position list probability outcome?
nine candidates for hire base on gender, six men and three women with only two position what is the list of probabilily outcome

Answer 1

You have three possiblities:
1. You hire two men
Probability = (6/9)*(5/8) = 5/12

2. You hire two women
Probability = (3/9)*(2/8) = 1/12

3. You hire one man and one women
Probability = (6/9)*(3/8) = 1/4

Hope this is what you are looking for, and if not, oh well. :)

Answer 2

I assume you are supposed to fill the positions at random.

There are 36 possible ways you can choose two people from a group of nine.

There are 15 different ways that you can pick two men. So, the probability of picking two men is 15/36 = 0.41666667

There are three different ways that you can choose two women. So, the probability of picking two women is 3/36 = 0.08333333.

That leaves 18 ways that you can choose one man and one woman -- so the probability of picking one of each is 0.5.
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Note -- Smiley Girl forgot that the order doesn't matter. She got 1/4 for the probability of hiring a man and a woman because that is the probability of hiring a man first then a woman from the remaining group. You also have the possibility of hiring a woman first and then a man -- which gives you the other 1/4.

Answer 3

Depends totaly on suitability of each candidate for the two positions you couldnt really have a man as a female cloakroom assistant now could you(or could you!!!)