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yashar806Best ResponseYou've already chosen the best response.0
There are four patients on the neonatal 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

satellite73Best ResponseYou've already chosen the best response.0
no time interval is mentioned, so my only guess is to solve having 3 or 4 patients need help
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
(A) 0.0756 (B) 0.1104 (C) 0.0837 (D) 0.0463 (E) 0.2646
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
you are told that these are independent events, which is a set up for using the binomial probability
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
it is a multiple choice question
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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

satellite73Best ResponseYou've already chosen the best response.0
of course \(\binom{4}{3}=4\)
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
you mean solve for that equation
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=4%28.3%29^3*.7%2B.3^4
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
ok, i got it thanks so much
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
it is not an equation, it is a computation requiring a calculator for sure
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
sorry, where did you get that 3 ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
you mean when i wrote \(P(X=3)\) ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i was calculating the probability that 3 patients needed help
 one year ago

yashar806Best ResponseYou've already chosen the best response.0
why three of them needed help/?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
\[P(X=k)=\binom{n}{k}p^k(1p)^{nk}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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

yashar806Best ResponseYou've already chosen the best response.0
oh, now I totally got it
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
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