- 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

- katieb

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- misty1212

HI!!

- 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

I do not get that can you explain further please?

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## More answers

- anonymous

I do not know how to find that probability. That is what I am struggling with

- 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

the probability that two are in that same dorm room is \(\frac{1}{14}\times \frac{1}{14}\)

- misty1212

the probability that thre are in that same dorm room is \(\left(\frac{1}{14}\right)^3\)

- misty1212

and the probability that all four are is \(\left(\frac{1}{14}\right)^4\)

- misty1212

add those numbers up
that should give you the answer

- misty1212

\[\left(\frac{1}{14}\right)^2+\left(\frac{1}{14}\right)^3+\left(\frac{1}{14}\right)^4\]should do it

- anonymous

Uh that's not the answer

- anonymous

the answer is .37, but I do not know how to get to that answer.

- amistre64

i wonder if a binomial probability works here

- 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

nah, that doesnt get you the answer that you say is correct so its most likely a bad thought as well

- amistre64

how do we know the answer is .37?

- anonymous

that's why I am asking.

- 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

the teacher posted answers for us to check with

- 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

thats really the only one i can find using the google

- anonymous

ah forget about that problem
can you help me on another though?

- amistre64

maybe

- 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

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

the 15 choose 0 is not a good thought :) that would be 1, but we want none since 5 are already picked.

- amistre64

and those are prolly best as multiplications, not additions ... then it would work better

- amistre64

or brute it for proof
gggrr
gggrw
gggww
ggggr
ggggw
ggggg
as long as order doesnt matter

- amistre64

if order matters
gggrr , 10
gggrw , 20
gggww , 10
ggggr, 5
ggggw , 5
ggggg, 1

- anonymous

but the answer is 16002 according to the answers my teacher put up.

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