Question

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

Answer 1

HI!!

Answer 2

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

Answer 3

I do not get that can you explain further please?

Answer 4

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

Answer 5

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 \(\frac{1}{14}\)

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

add those numbers up
that should give you the answer

Answer 10

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

Answer 11

Uh that's not the answer

Answer 12

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

Answer 13

i wonder if a binomial probability works here

Answer 14

p = 1/14, q = 13/14

just trying to work out what the structure would look like if it was, maybe?
\[P(x)=\binom4xp^xq^{4-x}\]

Answer 15

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

Answer 16

how do we know the answer is .37?

Answer 17

that's why I am asking.

Answer 18

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?

Answer 19

the teacher posted answers for us to check with

Answer 20

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

Answer 21

thats really the only one i can find using the google

Answer 22

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

Answer 23

maybe

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

or brute it for proof

gggrr
gggrw
gggww

ggggr
ggggw

ggggg

as long as order doesnt matter

Answer 29

if order matters

gggrr , 10

gggrw , 20

gggww , 10

ggggr, 5
ggggw , 5

ggggg, 1

Answer 30

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