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anonymous

  • 5 years ago

Ten students are applying for 3 positions on a team. the students include 4 boys (adam, alex, anthony, and arnold) and 6 girls (abbey, aurora, agnes, alice, amanda, and anna). all the students have an equal chance of being selected. find the probability that the students selected will include: a) at most 1 girl b) adam, anthony, and alice c) agnes and 2 other students

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  1. anonymous
    • 5 years ago
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    ur in college?

  2. anonymous
    • 5 years ago
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    naww not in college lol :P highschool why'd u think im in college?

  3. M
    • 5 years ago
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    the first one should be (4/10)^3 + (6/10)^1 x (4/10)^2

  4. M
    • 5 years ago
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    do you have answers to confirm?

  5. anonymous
    • 5 years ago
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    i was working on his question until you answered it. i was simply wondering why theis person thought i was in college. he asked me a question, i was answering.

  6. M
    • 5 years ago
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    second one should be (6/10)^1 x (4/10)^2

  7. M
    • 5 years ago
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    is it with replacement or without? lol

  8. anonymous
    • 5 years ago
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    What do you think?

  9. M
    • 5 years ago
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    i guess without :P

  10. M
    • 5 years ago
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    yeah i always get confused with statistics

  11. M
    • 5 years ago
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    you may have to use hypergeometric forumula

  12. M
    • 5 years ago
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    wwhat i wrote is binomial

  13. anonymous
    • 5 years ago
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    Well, you certainly don't 'have to' - if one really wanted to they could solve this by just drawing a (massive) tree diagram, but I think we can be slightly more clever about it,

  14. M
    • 5 years ago
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    hmm maybe (6/10)(4/9)(3/8) x 3 + (4/10)(3/9)(2/8)?

  15. M
    • 5 years ago
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    crap i'm so confused with this i should just stop lol. gl with the problem.

  16. anonymous
    • 5 years ago
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    I don't have answers to go with this.

  17. anonymous
    • 5 years ago
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    anyone figure this out?

  18. anonymous
    • 5 years ago
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    Don't worry, you can just draw a tree-diagram with 720 branches and be sure of the answer.

  19. M
    • 5 years ago
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    [(6/10)(4/9)(3/8) + (4/10)(6/9)(3/8) + (4/10)(3/9)(6/8)] + (4/10)(3/9)(2/8)

  20. anonymous
    • 5 years ago
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    Proof?

  21. M
    • 5 years ago
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    wait the ordering doesn't matter so you have to divide the first part by 3? or 6?

  22. anonymous
    • 5 years ago
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    I preferred it in the old form you posted: (6/10)(4/9)(3/8) x 3 + (4/10)(3/9)(2/8)

  23. M
    • 5 years ago
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    only problem with that is you don't know girl is selected before boys

  24. anonymous
    • 5 years ago
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    You don't. But your answer there accounts for it by multiplying by three.

  25. M
    • 5 years ago
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    i don't know stat well enough to figure this out without final answer unfortunatley

  26. anonymous
    • 5 years ago
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    OK. If there are three boys chosen, the chance is (clearly... I hope) (4/10)(3/9)(2/8) The only other option is 1 girl and 2 boys. This can either be GBB, BGB, BBG, but the chance of each is still (6x4x3)/(10x8x9). So your answer \[\frac{(6 \times 4 \times 3) \times 3 + 4 \times 3 \times 2}{720} \] is right.

  27. anonymous
    • 5 years ago
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    for which part?

  28. M
    • 5 years ago
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    even though probability of getting boy or girl is different?

  29. M
    • 5 years ago
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    oh wait you already accounted for that lol

  30. anonymous
    • 5 years ago
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    Yes; there's still a total number of 10*9*8 = 720 in the denominator; the order of the multiplication above is irrelevant. @ azntiger627 - FACEPALM - if you honestly have no idea which part this is then you have clearly done nothing on this work, and do not deserve help until at least attempt it.

  31. anonymous
    • 5 years ago
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    I just came back from dinner lol

  32. anonymous
    • 5 years ago
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    Wasting time eating... when you have Maths to do!? That's a cardinal sin in my book.

  33. anonymous
    • 5 years ago
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    lol sorry, now I'm rdy, so which part were u guys talking about?

  34. anonymous
    • 5 years ago
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    I'll tell you. But you have to guess first.

  35. anonymous
    • 5 years ago
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    a?

  36. anonymous
    • 5 years ago
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    Good job.

  37. anonymous
    • 5 years ago
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    ok so what did u find out so far?

  38. anonymous
    • 5 years ago
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    Nothing :(

  39. M
    • 5 years ago
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    Newton already solved part (a) for you above. but i think you should go over permutations and combinations before trying to solve the question above. good luck!

  40. anonymous
    • 5 years ago
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    You solved it first :(

  41. anonymous
    • 5 years ago
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    D) ?

  42. anonymous
    • 5 years ago
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    for d, i'm guessing (4C2)(6C1) / 10C3

  43. anonymous
    • 5 years ago
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    Very funny, Mr; there is no D.

  44. anonymous
    • 5 years ago
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    lol b I meant

  45. anonymous
    • 5 years ago
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    is that correct though?

  46. anonymous
    • 5 years ago
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    That equal 3/10 - does that seem correct?

  47. anonymous
    • 5 years ago
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    yea I got 3/10, is it correct though?

  48. anonymous
    • 5 years ago
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    There are 720 possibilities, and you think 216 contain the same three people? (they don't)

  49. anonymous
    • 5 years ago
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    what do u think then?

  50. anonymous
    • 5 years ago
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    I don't 'think' anything; I know what the answer is.

  51. anonymous
    • 5 years ago
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    what is it

  52. anonymous
    • 5 years ago
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    i will look at it soon, im tied up atm

  53. anonymous
    • 5 years ago
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    There names all begin with A, which is annoying, so just called them ABC. There are 6 ways of arranging them: ABC ACB etc. And 10 x 9 x 8 possible ways in total. Continue from there.

  54. anonymous
    • 5 years ago
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    Their names* UGGGGGGGGGH grammar fail.

  55. anonymous
    • 5 years ago
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    6/720?

  56. anonymous
    • 5 years ago
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    ¬_¬ maybe

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