Question

Probability

Answer 1

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?

Answer 2

no time interval is mentioned, so my only guess is to solve having 3 or 4 patients need help

Answer 3

(A) 0.0756 (B) 0.1104 (C) 0.0837 (D) 0.0463 (E) 0.2646

Answer 4

you are told that these are independent events, which is a set up for using the binomial probability

Answer 5

it is a multiple choice question

Answer 6

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

Answer 7

of course \(\binom{4}{3}=4\)

Answer 8

you mean solve for that equation

Answer 9

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

Answer 10

ok, i got it thanks so much

Answer 11

it is not an equation, it is a computation
requiring a calculator for sure

Answer 12

yw

Answer 13

sorry, where did you get that 3 ?

Answer 14

you mean when i wrote \(P(X=3)\) ?

Answer 15

is int 2?

Answer 16

yes

Answer 17

i was calculating the probability that 3 patients needed help

Answer 18

how.?

Answer 19

why three of them needed help/?

Answer 20

\[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\]

Answer 21

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

Answer 22

oh, now I totally got it

Answer 23

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?"

Answer 24

ok thx

Answer 25

yw