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There are five empty chairs in a row. If six men and four women are waiting to be seated, what is the probability that the seats will be occupied by two men and three women?
Correct Answer: B
The probability of seating two men and three women in the empty chairs = (total number of ways of choosing 2 men out of 6 and 3 women out of 4)/ (number of ways of seating five people out of ten).
  1. This is a combination-probability problem. To solve this we will need to find out the number of ways we can choose 2 men out of 6 and 3 women out of 4 to sit in the chair.
  2. Next, we note that the probability of seating two men and three women in the empty chairs is:
    (number of ways of choosing 2 men and 3 women)/ (number of ways of seating five people out of ten).
  3. The total number of ways of choosing five people to sit out of a total of ten (six men + four women) is given by:
    10C5 = 10!/([10-5]!5!)
    = (10*9*8*7*6)/[10-5]!
    = (10*9*8*7*6)/5!
    = 252
  4. Now we want the seats to be taken by 2 men and 3 women.
    I. The number of ways to choose 2 men out of six to occupy the seat is given by:
    6C2 = 6!/4!*2! = 15
    II. The number of ways to choose 3 women out of 4 to sit is given by:
    4C3 = 4!/3!*1! = 4
  5. Using the multiplication principle, the total number of ways to select two men and three women is:
    15*4=60
  6. The probability is thus = 60/252
    = (12*5)/(12*21)
    = 5/21. Hence the correct answer choice is B

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