## 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

1. misty1212

HI!!

2. 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

3. anonymous

I do not get that can you explain further please?

4. anonymous

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

5. 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 $$\frac{1}{14}$$

6. misty1212

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

7. misty1212

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

8. misty1212

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

9. misty1212

10. misty1212

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

11. anonymous

12. anonymous

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

13. amistre64

i wonder if a binomial probability works here

14. amistre64

p = 1/14, q = 13/14 just trying to work out what the structure would look like if it was, maybe? $P(x)=\binom4xp^xq^{4-x}$

15. amistre64

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

16. amistre64

how do we know the answer is .37?

17. anonymous

18. 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?

19. anonymous

the teacher posted answers for us to check with

20. 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

21. amistre64

thats really the only one i can find using the google

22. anonymous

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

23. amistre64

maybe

24. 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.

25. 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

26. amistre64

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

27. amistre64

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

28. amistre64

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

29. amistre64

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

30. anonymous

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