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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?
no time interval is mentioned, so my only guess is to solve having 3 or 4 patients need help
(A) 0.0756 (B) 0.1104 (C) 0.0837 (D) 0.0463 (E) 0.2646

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Other answers:

you are told that these are independent events, which is a set up for using the binomial probability
it is a multiple choice question
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
of course \(\binom{4}{3}=4\)
you mean solve for that equation
http://www.wolframalpha.com/input/?i=4%28.3%29^3*.7%2B.3^4
ok, i got it thanks so much
it is not an equation, it is a computation requiring a calculator for sure
yw
sorry, where did you get that 3 ?
you mean when i wrote \(P(X=3)\) ?
is int 2?
yes
i was calculating the probability that 3 patients needed help
how.?
why three of them needed help/?
\[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\]
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
oh, now I totally got it
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?"
ok thx
yw

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