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yashar806

  • one year ago

Probability

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  1. yashar806
    • one year ago
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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?

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

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

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

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

  6. satellite73
    • one year ago
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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

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

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

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

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

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

  12. satellite73
    • one year ago
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    yw

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

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

  15. yashar806
    • one year ago
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    is int 2?

  16. yashar806
    • one year ago
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    yes

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

  18. yashar806
    • one year ago
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    how.?

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

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

  21. satellite73
    • one year ago
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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

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

  23. satellite73
    • one year ago
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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?"

  24. yashar806
    • one year ago
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    ok thx

  25. satellite73
    • one year ago
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    yw

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