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yashar806

Probability

  • one year ago
  • one year ago

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  1. yashar806
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    There are four patients on the neo-natal ward of a local hospital who are monitored by two staff members. Suppose the probability (at any one time) of a patient requiring attention by a sta member is 0.3. Assuming the patients behave independently, what is the probability at any one time that there will not be suffccient staff to attend to all patients who need them?

    • one year ago
  2. satellite73
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    no time interval is mentioned, so my only guess is to solve having 3 or 4 patients need help

    • one year ago
  3. yashar806
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    (A) 0.0756 (B) 0.1104 (C) 0.0837 (D) 0.0463 (E) 0.2646

    • one year ago
  4. satellite73
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    you are told that these are independent events, which is a set up for using the binomial probability

    • one year ago
  5. yashar806
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    it is a multiple choice question

    • one year ago
  6. satellite73
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    the probability that three will need it is \[P(x=3)=\binom{4}{3}(.3)^3(.7)\] and the probability that all four will need it is \[P(x=4)=.3^4\] add these up

    • one year ago
  7. satellite73
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    of course \(\binom{4}{3}=4\)

    • one year ago
  8. yashar806
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    you mean solve for that equation

    • one year ago
  9. satellite73
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    http://www.wolframalpha.com/input/?i=4%28.3%29^3*.7%2B.3^4

    • one year ago
  10. yashar806
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    ok, i got it thanks so much

    • one year ago
  11. satellite73
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    it is not an equation, it is a computation requiring a calculator for sure

    • one year ago
  12. satellite73
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    yw

    • one year ago
  13. yashar806
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    sorry, where did you get that 3 ?

    • one year ago
  14. satellite73
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    you mean when i wrote \(P(X=3)\) ?

    • one year ago
  15. yashar806
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    is int 2?

    • one year ago
  16. yashar806
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    yes

    • one year ago
  17. satellite73
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    i was calculating the probability that 3 patients needed help

    • one year ago
  18. yashar806
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    how.?

    • one year ago
  19. yashar806
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    why three of them needed help/?

    • one year ago
  20. satellite73
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    \[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\]

    • one year ago
  21. satellite73
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    there are two staff members and 4 patients if there are not enough staff members that means either 3 patients need help or 4 patients need help

    • one year ago
  22. yashar806
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    oh, now I totally got it

    • one year ago
  23. satellite73
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    the question asked "what is the probability at any one time that there will not be suffccient staff to attend to all patients who need them?"

    • one year ago
  24. yashar806
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    ok thx

    • one year ago
  25. satellite73
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    yw

    • one year ago
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