anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
5 choose 3 divided by 21 choose 3
anonymous
  • anonymous
so 3?
anonymous
  • anonymous
is it 5/26?

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anonymous
  • anonymous
no
anonymous
  • anonymous
Satellite already gave you the answer.
anonymous
  • anonymous
is it 3?
anonymous
  • anonymous
whats does 5 choose 3 devided by 21 choose 3 mean its only suppose to be 1 number rite
anonymous
  • anonymous
Choose is the name we sometimes give to the binomial coefficient function: \[\text{n choose k} = {n\choose k} = \frac{n!}{k!(n-k)!}\]
anonymous
  • anonymous
5!/3(21-3)!
anonymous
  • anonymous
Not quite
anonymous
  • anonymous
5 choose 3 = \(\frac{5!}{3!(5-3)!}\) 21 choose 3 = \(\frac{21!}{3!(21-3)!}\) Then divide those two values to find \[\frac{\text{5 choose 3}}{\text{21 choose 3}}\]
anonymous
  • anonymous
18-7
anonymous
  • anonymous
no
anonymous
  • anonymous
because when u do 21/3 it equals 7
anonymous
  • anonymous
Are you being taught to do this on your calculator using the Cr function?
anonymous
  • anonymous
no we don't use our calculators for it
anonymous
  • anonymous
Doing it by hand you have \[\frac{21!}{3!*18!} = \frac{21*20*19*18!}{3! * 18!} = \frac{21*20*19}{3*2} \]\[= 7*10*19 = 70*19 = 700 + 630= 1330\]
anonymous
  • anonymous
But that will make for a very long evening.
anonymous
  • anonymous
can i butt in for a moment?
anonymous
  • anonymous
it as 1/260
anonymous
  • anonymous
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
anonymous
  • anonymous
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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