Bankteller Problem: How Many Ways Can Both Men and Women Be Served?

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The Bankteller Problem involves calculating the number of ways to serve customers at a bank, specifically focusing on combinations of men and women. For the first scenario, the correct calculation for selecting 4 people, including 2 men and 2 women, is (6 choose 2) * (4 choose 2), resulting in 90 ways. The probability of randomly selecting 2 men and 2 women from a total of 10 people is calculated as (6C2 * 4C2) / (10C4), yielding a probability of 3/7. In the second scenario, for selecting 5 people with 3 men and 2 women, the calculation involves (10C5) * (6C3) * (4C2). The probability of this selection is determined similarly, resulting in 10/21.
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HELP! Bankteller Problem

1. Homework Statement
There are 6 males and 4 females awaiting to see a teller at a bank.

Only 4 people can be served at one time.
1) How many ways can four of the people be picked and served one at a time, if they must include two(2) men and two(2) women?


2) If indeed the four people are picked randomly, what is the probability that the four will include two (2) men and two (2) women?
This is the question I am confused about.


3. The Attempt at a Solution
My solution for Problem#1: (6 choose 2) * (4 choose 2) = 90
 
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2) 6C2 x 4C2 /10C4 = 3/7
 
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HELP! Bankteller Problem

Thank you. Could you check this.

It is the end of the day and there are 5 tellers, but each teller can only serve one person. Therefore only five people can be served, each by a teller, so the five people are picked at once.

1) How many ways can five of the people picked and served, if they must include 3 men and 2 women?

My solution: (10c5)*(6c3)*(4c2)

2) If indeed the 5 people are picked randomly, what is the probability that the five will include 3 men and 2 women?

My solution: If the solution to the first question is correct I should be able to figure it out.
 
This is bound to be homework.
 
Actually review of an even answer in the text.
 
2) Just as before,
6C3 x 4C2 / 10C5 =10/21
 
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