Question 1192754: A team of five is to be chosen from four men and five women to work on a special project
i. In how many ways can the team be chosen?
ii. In how many ways can the team be chosen to include just three women?
iii. What is the probability that the team just includes three women?
iv. What is the probability that the team include at least three women?
v. What is the probability that the team includes more men than women
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 9C5 ways to choose the groups=126 ways
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ii. numerator its 5C3*4C2 for 3 women and therefore 2 men=60, so there are 60 ways
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iii probability is 60/126 or 10/21.
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iv. at least 3 women includes 3,4,5. We know 60 ways for 3. For 4 it is 5C4*4C1=20 ways and for 5 women there is only one way. Therefore , there is 81/126 probability or 9/14.
v.Look at 2 women this has 5C2*4C3=40 ways
For 1 woman 4C4*5C1=5 ways.
Can't be 0 women.
1woman-4 men=5 ways
2W 3M-40 ways
3W2M-60 ways
4W1M-20 ways
5W0M-1 way
They add up to 126. They have to.
More men then women occurs 4/1 (5 ways) 3/2 (40 ways) no other ways. so probability is 45/126=15/34
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