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anonymous

  • 4 years ago

Assume that there are 8 different issues of Popular Science, 7 different issues of Time, and 4 different issues of Sports Illustrated, including the December 1st issue, on the rack. You choose 4 of them at random. (1) What is the probability that you choose 3 issues of Popular Science and 1 issue of Time? (2) What is the probability that you choose at least 3 of the Time magazines?

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  1. dumbcow
    • 4 years ago
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    P(1) = 4C3 * (7/19)*(8/18)*(7/17)*(6/16)

  2. Directrix
    • 4 years ago
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    P(3 PS and 1 Ti) = [C(8,3)C(7,1)] / C(19,4) = 98/969 = .101 approx

  3. anonymous
    • 4 years ago
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    no

  4. anonymous
    • 4 years ago
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    that didnt work

  5. Directrix
    • 4 years ago
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    See if you wrote the problem correctly. I was expecting a restriction about the Dec 1 SI.

  6. Directrix
    • 4 years ago
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    Mr. Cow has 98/969 = .101 approx

  7. anonymous
    • 4 years ago
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    it never specifies the 1 december 1st issue

  8. Directrix
    • 4 years ago
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    On the first part, Mr. Cow and I agree. Check every word of the problem.

  9. anonymous
    • 4 years ago
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    98/969 is correct .101 is not

  10. anonymous
    • 4 years ago
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    same thing right

  11. Directrix
    • 4 years ago
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    yes, are you to leave answers as fractions?

  12. anonymous
    • 4 years ago
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    i guess so it only worked with 969 but says .101 i tried it multiple times haha

  13. anonymous
    • 4 years ago
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    the next one does specify

  14. dumbcow
    • 4 years ago
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    for next part -> .1174 or 455/3876

  15. anonymous
    • 4 years ago
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    Genius guys

  16. Directrix
    • 4 years ago
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    2) at least 3 T [C(7,3) C(8,1) + C(7,3)C(4,1) +C(7,4)] / C(19,4) = 455 / 3 876 = .117 approx

  17. Directrix
    • 4 years ago
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    You are a genius guy. What's up with my fraction. I'll check.

  18. anonymous
    • 4 years ago
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    thank you so much man you have helped save me

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