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yashar806

  • 2 years ago

Probability

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  1. yashar806
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    (A) 0.0756 (B) 0.1104 (C) 0.0837 (D) 0.0463 (E) 0.2646

  4. satellite73
    • 2 years 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
    • 2 years ago
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    it is a multiple choice question

  6. satellite73
    • 2 years 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
    • 2 years ago
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    of course \(\binom{4}{3}=4\)

  8. yashar806
    • 2 years ago
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    you mean solve for that equation

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

  10. yashar806
    • 2 years ago
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    ok, i got it thanks so much

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

  12. satellite73
    • 2 years ago
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    yw

  13. yashar806
    • 2 years ago
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    sorry, where did you get that 3 ?

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

  15. yashar806
    • 2 years ago
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    is int 2?

  16. yashar806
    • 2 years ago
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    yes

  17. satellite73
    • 2 years ago
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    i was calculating the probability that 3 patients needed help

  18. yashar806
    • 2 years ago
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    how.?

  19. yashar806
    • 2 years ago
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    why three of them needed help/?

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

  21. satellite73
    • 2 years 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
    • 2 years ago
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    oh, now I totally got it

  23. satellite73
    • 2 years 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
    • 2 years ago
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    ok thx

  25. satellite73
    • 2 years ago
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    yw

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