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anonymous

  • 4 years ago

7 females, and 5 males including Larry. There are 4 tasks to be assigned. Note that assigning the same people different tasks constitutes a different assignment. (1) Find the probability that both males and females are given a task. (2) Find the probability that Larry and at least one female are given tasks.

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  1. Directrix
    • 4 years ago
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    Hey, CW.

  2. anonymous
    • 4 years ago
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    hey

  3. anonymous
    • 4 years ago
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    wow check this out http://www.jiskha.com/display.cgi?id=1236051521

  4. anonymous
    • 4 years ago
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    larry must be a popular guy

  5. anonymous
    • 4 years ago
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    how ya doin g

  6. anonymous
    • 4 years ago
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    doing

  7. anonymous
    • 4 years ago
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    these word problems are killing me

  8. anonymous
    • 4 years ago
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    it is the same problem with different numbers

  9. anonymous
    • 4 years ago
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    I had to take an exam today. And this homework set is due on tomorrow at 730 am. :)

  10. anonymous
    • 4 years ago
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    2 days to complete them

  11. anonymous
    • 4 years ago
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    no i am not....

  12. anonymous
    • 4 years ago
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    I like to know what i am doing though

  13. Directrix
    • 4 years ago
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    Wyatt Earp: Speed is great but accuracy is everything.

  14. anonymous
    • 4 years ago
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    exactly

  15. anonymous
    • 4 years ago
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    i am always mixing these up as I work through them

  16. anonymous
    • 4 years ago
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    I have 12 or 13 problems due in the morning

  17. anonymous
    • 4 years ago
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    I feel like a math bum :)

  18. Directrix
    • 4 years ago
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    Don't feel. Just work. Post the other 12 or 13 problems individually on this site. You do that while I work on this problem. I may have to stop and eat some brain food. :} You need to focus. Nobody here is a bum.

  19. anonymous
    • 4 years ago
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    is that the final solution

  20. anonymous
    • 4 years ago
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    for both male and female

  21. anonymous
    • 4 years ago
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    i know I am very lost haha

  22. anonymous
    • 4 years ago
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    it didnt work though

  23. anonymous
    • 4 years ago
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    why the zero in both combinations

  24. anonymous
    • 4 years ago
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    not quite brain food :) and this one

  25. anonymous
    • 4 years ago
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    ha ha i post them wuickly

  26. anonymous
    • 4 years ago
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    quickly

  27. Directrix
    • 4 years ago
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    P(both genders) = 1 - { [ C(7,0) C(5,4) - C(7,4) C(5,0)] / C(12,4) } = 1 - 40/495 = 455/495 = 91/99 = .919 approx

  28. anonymous
    • 4 years ago
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    Correct

  29. anonymous
    • 4 years ago
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    So the second part would be similar?

  30. Directrix
    • 4 years ago
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    P(Larry and at least 1 F) P(at least 1F) = 1 - P(0F) = 1 -{ [C(7,0)C(5,4)] } / C(12,4) = 5/495 = 490/495 = 98/99 = .99 ish. Ways to choose Larry = 1 C(1,1) = 1 P(Larry and at least 1 F) = 1(98/99) = 98/99 = .99 approx

  31. anonymous
    • 4 years ago
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    no it didnt work

  32. anonymous
    • 4 years ago
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    larry wouldnt be 5?

  33. anonymous
    • 4 years ago
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    5,1

  34. Directrix
    • 4 years ago
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    But, we want Larry and none of the other males. So P(Larry) = C(1,1) C(4,0) = 1. That would be Larry and none of the other males.

  35. Directrix
    • 4 years ago
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    I'm going to cases for this. Slow but may make more sense.

  36. anonymous
    • 4 years ago
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    how ya doing

  37. Directrix
    • 4 years ago
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    Larry and at least 1F -------- Cases {Lar, 1F, 2 other M}; {Lar, 2F, 1 other M}; and {Lar, 3F, 0 other M}

  38. Directrix
    • 4 years ago
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    Ways to choose Lar, 1F, 2 other M} = 1 C(7,1) C(4,2) = 42 Ways to choose {Lar, 2F, 1 other M}; = 1 C(7,2) C(4,1)= 84 Ways to choose {Lar, 3F, 0 other M} = 1 C(7,3) C(4,0) = 35 Ways to choose 4 from 12 is C(12,4) = 495 P(L and at least 1F) = [42 + 84 +35] / 495 = 161/495 =.325 approx

  39. anonymous
    • 4 years ago
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    CORRECT! You are a genius.... I have posted some more. I haven't wanted to post to many at a time and flood the screen

  40. anonymous
    • 4 years ago
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    I am wrong no matter what i do on these

  41. Directrix
    • 4 years ago
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    Let's find another problem.

  42. anonymous
    • 4 years ago
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    I used a lot of the work i learned from hear that helped me on the test in some of these.

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