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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

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    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
    • one year ago
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    I do not get that can you explain further please?

  4. anonymous
    • one year ago
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    I do not know how to find that probability. That is what I am struggling with

  5. misty1212
    • one year ago
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    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
    • one year ago
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    the probability that two are in that same dorm room is \(\frac{1}{14}\times \frac{1}{14}\)

  7. misty1212
    • one year ago
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    the probability that thre are in that same dorm room is \(\left(\frac{1}{14}\right)^3\)

  8. misty1212
    • one year ago
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    and the probability that all four are is \(\left(\frac{1}{14}\right)^4\)

  9. misty1212
    • one year ago
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    add those numbers up that should give you the answer

  10. misty1212
    • one year ago
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    \[\left(\frac{1}{14}\right)^2+\left(\frac{1}{14}\right)^3+\left(\frac{1}{14}\right)^4\]should do it

  11. anonymous
    • one year ago
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    Uh that's not the answer

  12. anonymous
    • one year ago
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    the answer is .37, but I do not know how to get to that answer.

  13. amistre64
    • one year ago
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    i wonder if a binomial probability works here

  14. amistre64
    • one year ago
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    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
    • one year ago
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    nah, that doesnt get you the answer that you say is correct so its most likely a bad thought as well

  16. amistre64
    • one year ago
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    how do we know the answer is .37?

  17. anonymous
    • one year ago
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    that's why I am asking.

  18. amistre64
    • one year ago
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    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
    • one year ago
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    the teacher posted answers for us to check with

  20. amistre64
    • one year ago
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    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
    • one year ago
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    thats really the only one i can find using the google

  22. anonymous
    • one year ago
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    ah forget about that problem can you help me on another though?

  23. amistre64
    • one year ago
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    maybe

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

  27. amistre64
    • one year ago
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    and those are prolly best as multiplications, not additions ... then it would work better

  28. amistre64
    • one year ago
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    or brute it for proof gggrr gggrw gggww ggggr ggggw ggggg as long as order doesnt matter

  29. amistre64
    • one year ago
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    if order matters gggrr , 10 gggrw , 20 gggww , 10 ggggr, 5 ggggw , 5 ggggg, 1

  30. anonymous
    • one year ago
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    but the answer is 16002 according to the answers my teacher put up.

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