There are 10 patients on the neonatal ward of a local hospital who are monitored by 2 staff members. If the probability (at any one time) of a patient requiring emergency attention by a staff member is .3, assuming the patients to be behave independently, what is the probability at any one time that there will not be sufficient staff to attend all emergencies?
Solution
This is a binomial distribution with parameters
n = 10, p = 0.3 and q = 0.7
the formula for k patients simultaneously needing attention will
be
P[k] = 10Ck*(0.3)k*(0.7)(10-k)
we want to find P(k[>2) = 1 - (P[0] + P[1] +P[2])
= 1 - [ (0.7)10 + 10C1*(0.3)(0.7)9 +
10C2(0.3)2(0.7)8 ]
= 0.6172
There are 10 patients on the neonatal ward of a local hospital who are monitored by...
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