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anonymous

  • 5 years ago

assume that a perfessional baseball team has 11 ptchers,5 infielders, and 10 other players. if 3 players names are selected at random, determine the probl that all 3 are infielders.

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  1. anonymous
    • 5 years ago
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    5 choose 3 divided by 21 choose 3

  2. anonymous
    • 5 years ago
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    so 3?

  3. anonymous
    • 5 years ago
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    is it 5/26?

  4. anonymous
    • 5 years ago
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    no

  5. anonymous
    • 5 years ago
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    Satellite already gave you the answer.

  6. anonymous
    • 5 years ago
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    is it 3?

  7. anonymous
    • 5 years ago
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    whats does 5 choose 3 devided by 21 choose 3 mean its only suppose to be 1 number rite

  8. anonymous
    • 5 years ago
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    Choose is the name we sometimes give to the binomial coefficient function: \[\text{n choose k} = {n\choose k} = \frac{n!}{k!(n-k)!}\]

  9. anonymous
    • 5 years ago
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    5!/3(21-3)!

  10. anonymous
    • 5 years ago
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    Not quite

  11. anonymous
    • 5 years ago
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    5 choose 3 = \(\frac{5!}{3!(5-3)!}\) 21 choose 3 = \(\frac{21!}{3!(21-3)!}\) Then divide those two values to find \[\frac{\text{5 choose 3}}{\text{21 choose 3}}\]

  12. anonymous
    • 5 years ago
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    18-7

  13. anonymous
    • 5 years ago
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    no

  14. anonymous
    • 5 years ago
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    because when u do 21/3 it equals 7

  15. anonymous
    • 5 years ago
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    Are you being taught to do this on your calculator using the Cr function?

  16. anonymous
    • 5 years ago
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    no we don't use our calculators for it

  17. anonymous
    • 5 years ago
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    Doing it by hand you have \[\frac{21!}{3!*18!} = \frac{21*20*19*18!}{3! * 18!} = \frac{21*20*19}{3*2} \]\[= 7*10*19 = 70*19 = 700 + 630= 1330\]

  18. anonymous
    • 5 years ago
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    But that will make for a very long evening.

  19. anonymous
    • 5 years ago
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    can i butt in for a moment?

  20. anonymous
    • 5 years ago
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    it as 1/260

  21. anonymous
    • 5 years ago
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    If i want to know the number of ways to choose 3 people (items, whatever) out of a set of 21 I reason as follows: there are 21 choices for the first person, 20 for the next and then 19 for the last. by the counting principle there are then 21*20*19 ways to do this. But this counts (a,b,c) differently than (b,c,a) for example so I have over-counted by the number of ways I can permute 3 things. There are 3*2 = 6 such permutations. Thus the number of ways to choose 3 from a set of 21 is 21*20*19/3*2. Similarly the number of ways to choose 3 from a set of 5 is 5*4*3/3*2 and the number of ways to choose 4 out of 11 is 11*10*9*8/4*3*2 etc

  22. anonymous
    • 5 years ago
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    Hello, Your chance of picking one is 5/26, then, once you've picked him, there are only 4 left out of 25 and once you've picked that player, there's only 3 left out of 24 to choose from (like in elementary school picking teams) giving you: (5/26)*(4/25)*(3/24) = 1/260, like you said.

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