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anonymous
 one year ago
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
anonymous
 one year ago
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

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.2at 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
 one year ago
Best ResponseYou've already chosen the best response.0I do not get that can you explain further please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do not know how to find that probability. That is what I am struggling with

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2maybe 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
 one year ago
Best ResponseYou've already chosen the best response.2the probability that two are in that same dorm room is \(\frac{1}{14}\times \frac{1}{14}\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2the probability that thre are in that same dorm room is \(\left(\frac{1}{14}\right)^3\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2and the probability that all four are is \(\left(\frac{1}{14}\right)^4\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2add those numbers up that should give you the answer

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2\[\left(\frac{1}{14}\right)^2+\left(\frac{1}{14}\right)^3+\left(\frac{1}{14}\right)^4\]should do it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Uh that's not the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer is .37, but I do not know how to get to that answer.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i wonder if a binomial probability works here

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0p = 1/14, q = 13/14 just trying to work out what the structure would look like if it was, maybe? \[P(x)=\binom4xp^xq^{4x}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0nah, that doesnt get you the answer that you say is correct so its most likely a bad thought as well

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0how do we know the answer is .37?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's why I am asking.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0im 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
 one year ago
Best ResponseYou've already chosen the best response.0the teacher posted answers for us to check with

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0http://www.algebra.com/algebra/homework/Probabilityandstatistics/Probabilityandstatistics.faq.question.339416.html this one gets the same results that i did ... using binomial ill check others

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0thats really the only one i can find using the google

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah forget about that problem can you help me on another though?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0An 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
 one year ago
Best ResponseYou've already chosen the best response.0does 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
 one year ago
Best ResponseYou've already chosen the best response.0the 15 choose 0 is not a good thought :) that would be 1, but we want none since 5 are already picked.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0and those are prolly best as multiplications, not additions ... then it would work better

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0or brute it for proof gggrr gggrw gggww ggggr ggggw ggggg as long as order doesnt matter

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0if order matters gggrr , 10 gggrw , 20 gggww , 10 ggggr, 5 ggggw , 5 ggggg, 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the answer is 16002 according to the answers my teacher put up.
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