Question 867569: A task force studying sex discrimination wishes to establish a subcommittee. If the subcommittee is to consist of 11 members chosen at random from a group of 19 men and 19 women, find the probability that the subcommittee consists of the following?
(a) 6 men and 5 women (Round your answer to two decimal places.)
(b) all women (Round your answer to seven decimal places.)
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! There are 19+19 = 38 people total. So there are 38 C 11 = 1,203,322,288 ways to form a committee of 11 people (with no restrictions on gender).
I'm using the combination formula to compute 38 C 11. It's faster to do with a calculator, but I recommend you do a few by hand to get an idea of how it works.
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a)
There are 19 C 6 = 27,132 ways to pick 6 men (from a pool of 19 men)
There are 19 C 5 = 11,628 ways to pick 5 women (from a pool of 19 women)
There are 27,132*11,628 = 315,490,896 ways to pick a committee of 11 people where there are 6 men and 5 women.
Divide this by the total number of ways to form an 11 member committee: 315,490,896/1,203,322,288 = 0.26218320656586
Then round to two decimal places: 0.26218320656586 ----> 0.26
So the probability of selecting exactly 6 men and 5 women is 0.26
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b)
There are 19 C 11 = 75,582 ways to pick a committee of all women.
Divide this by the total number of ways to form an 11 member committee: 75,582/1,203,322,288 = 0.00006281110286
Then round to seven decimal places: 0.00006281110286 ----> 0.0000628
So the probability of selecting all women, ie a committee of 11 women, is 0.0000628
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A task force studying sex discrimination wishes to establish a subcommittee. If the subcommittee is to consist of 11 members chosen at random from a group of 19 men and 19 women, find the probability that the subcommittee consists of the following?
(a) 6 men and 5 women (Round your answer to two decimal places.
Ans: [19C6*19C5]/[38C11]
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(b) all women (Round your answer to seven decimal places.)
Ans: 19C11/38C11
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Cheers,
Stan H.
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