anonymous
  • anonymous
Assume that the committee consists of 8 Republicans and 5 Democrats. A subcommittee of 4 is randomly selected from all subcommittees of 4 which contain at least 1 Democrat. What is the probability that the new subcommittee will contain at least 2 Democrats?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
good lord
anonymous
  • anonymous
lets see if we can figure out how many subcommittees contain at least one democrat, and i guess the best way to do it is figure out how many contain no democrats and subtract it from the total possible subcommittees
anonymous
  • anonymous
these things ar ekilling me and i have to have these done by 730 am

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anonymous
  • anonymous
number of subcommittees with not restriction is \[\dbinom{13}{4}=\frac{13\times 12\times 11\times 10}{4\times 3\times 2}=715\]
anonymous
  • anonymous
number of subcommittees that are all republicans are \[\frac{8\times 7\times 6\times 5}{4!}=70\] so i guess we have \[715-70=645\] total subcommittees that contain at least one democrat
anonymous
  • anonymous
that will be our denominator. now we have to figure out, of these 645 subcommittees with at least one democrat, how many have at least two
anonymous
  • anonymous
ok exactly one democrat number of ways would be \[\dbinom{5}{1}\dbinom{8}{3}\] aka \[5\times 56=280\]
anonymous
  • anonymous
so subtract this off from 645 to get the number that contain more than 1 democrat. you get \[645-280=365\] so our answer is \[\frac{365}{645}\]
anonymous
  • anonymous
that didnt work
anonymous
  • anonymous
you sure? i think the method is right. maybe a calculation error?
anonymous
  • anonymous
the more i think about it the more i think it is right. do you have an answer?
anonymous
  • anonymous
you are right
anonymous
  • anonymous
whew!!
anonymous
  • anonymous
im sorry not sure why it didnt work first time
anonymous
  • anonymous
ok did you try the other one i sent? is this an on line class?
anonymous
  • anonymous
no but home work is submitted online...
anonymous
  • anonymous
oh i see. have fun!

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