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yashar806
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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?
anonymous
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no time interval is mentioned, so my only guess is to solve having 3 or 4 patients need help
yashar806
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(A) 0.0756 (B) 0.1104 (C) 0.0837 (D) 0.0463 (E) 0.2646
anonymous
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you are told that these are independent events, which is a set up for using the binomial probability
yashar806
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it is a multiple choice question
anonymous
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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
anonymous
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of course \(\binom{4}{3}=4\)
yashar806
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you mean solve for that equation
yashar806
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ok, i got it thanks so much
anonymous
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it is not an equation, it is a computation
requiring a calculator for sure
anonymous
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yw
yashar806
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sorry, where did you get that 3 ?
anonymous
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you mean when i wrote \(P(X=3)\) ?
yashar806
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is int 2?
yashar806
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yes
anonymous
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i was calculating the probability that 3 patients needed help
yashar806
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how.?
yashar806
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why three of them needed help/?
anonymous
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\[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\]
anonymous
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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
yashar806
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oh, now I totally got it
anonymous
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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?"
yashar806
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ok thx
anonymous
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yw