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anonymous
 3 years ago
Probability
anonymous
 3 years ago
Probability

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There 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?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(A) 0.0756 (B) 0.1104 (C) 0.0837 (D) 0.0463 (E) 0.2646

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is a multiple choice question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0of course \(\binom{4}{3}=4\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you mean solve for that equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, i got it thanks so much

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is not an equation, it is a computation requiring a calculator for sure

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry, where did you get that 3 ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i was calculating the probability that 3 patients needed help

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why three of them needed help/?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, now I totally got it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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?"
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