anonymous
  • anonymous
Four high school friends will all be attending the same university next year. There are 14 dormitories on campus. find the probability that at least 2 of the friends will be in the same dormitory
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
at least two in the same dorm is easiest to compute by computing the probability that they are all in different dorms, and subtracting that from 1
anonymous
  • anonymous
I do not get that can you explain further please?

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More answers

anonymous
  • anonymous
I do not know how to find that probability. That is what I am struggling with
misty1212
  • misty1212
maybe it is easier to do it directly he is in some dorm room. the probability that any one friend is in that same dorm room is \(\frac{1}{14}\)
misty1212
  • misty1212
the probability that two are in that same dorm room is \(\frac{1}{14}\times \frac{1}{14}\)
misty1212
  • misty1212
the probability that thre are in that same dorm room is \(\left(\frac{1}{14}\right)^3\)
misty1212
  • misty1212
and the probability that all four are is \(\left(\frac{1}{14}\right)^4\)
misty1212
  • misty1212
add those numbers up that should give you the answer
misty1212
  • misty1212
\[\left(\frac{1}{14}\right)^2+\left(\frac{1}{14}\right)^3+\left(\frac{1}{14}\right)^4\]should do it
anonymous
  • anonymous
Uh that's not the answer
anonymous
  • anonymous
the answer is .37, but I do not know how to get to that answer.
amistre64
  • amistre64
i wonder if a binomial probability works here
amistre64
  • amistre64
p = 1/14, q = 13/14 just trying to work out what the structure would look like if it was, maybe? \[P(x)=\binom4xp^xq^{4-x}\]
amistre64
  • amistre64
nah, that doesnt get you the answer that you say is correct so its most likely a bad thought as well
amistre64
  • amistre64
how do we know the answer is .37?
anonymous
  • anonymous
that's why I am asking.
amistre64
  • amistre64
im saying, is the answer given to you in the back of a book, or did you find someone solution online ... or how did you determine that it has to be .37 to start with?
anonymous
  • anonymous
the teacher posted answers for us to check with
amistre64
  • amistre64
http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.339416.html this one gets the same results that i did ... using binomial ill check others
amistre64
  • amistre64
thats really the only one i can find using the google
anonymous
  • anonymous
ah forget about that problem can you help me on another though?
amistre64
  • amistre64
maybe
anonymous
  • anonymous
An urn contains 10 red balls, 10 green balls, and 5 white balls. 5 balls are selected. In how many ways can 5 balls be drawn if at least 3 are green.
amistre64
  • amistre64
does this thought make sense? (10 choose 3) green + (15 choose 2) not green or (10 choose 4) green + (15 choose 1) not green or (10 choose 5) green + (15 choose 0) not green
amistre64
  • amistre64
the 15 choose 0 is not a good thought :) that would be 1, but we want none since 5 are already picked.
amistre64
  • amistre64
and those are prolly best as multiplications, not additions ... then it would work better
amistre64
  • amistre64
or brute it for proof gggrr gggrw gggww ggggr ggggw ggggg as long as order doesnt matter
amistre64
  • amistre64
if order matters gggrr , 10 gggrw , 20 gggww , 10 ggggr, 5 ggggw , 5 ggggg, 1
anonymous
  • anonymous
but the answer is 16002 according to the answers my teacher put up.

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