Question 1182763: Problem 5 (12%)
Suppose the number of admissions to the emergency room at a small hospital follows a Poisson distribution, but the incidence rate changes on different days of the week. On a weekday, there is on average two admission per day, while on a weekend day there is on average one admission per day.
a- What is the probability of at least one admission on a Wednesday?
b- What is the probability of at least two admissions on a Saturday?
c- What is the probability that in a given month having 4 Saturdays, there are exactly 3 Saturdays with at least two admissions each?
d- What is the probability of exactly 5 admissions in 2 weekdays?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! probability of 0 emissions on Wed. is p(0)=e^(-2)=0.1353
1-0.1353=0.8647, and that is the probability of at least one admission.
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on Saturday look at P(0) and P(1) for parameter 1; if we know that probability (1- that) will be at least 2.
for P(0), it is e^(-1)=0.3679
for P(1) it is e^(-1)*1^2/1=0.3679
That sum is 0.7358, so the answer is the complement or 1-0.7358=0.2642.
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probability of a Saturday with two admissions is 0.2642.
The answer will be 4C3*0.2642^3*0.7358=0.0543
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This isn't quite clear to me. Is it 5 admissions for 2 given independently selected weekdays?
If I look at a given two weekdays, and look for combinations that give 5,
They may be 0 and 5/5 and 0
1/4 and 4/1
3/2 and 2/3
P(0)=0.1353
p(1)=0.2707
p(2)=0.2707
p(3)=0.1804
p(4)=0.0902
p(5)=0.0361
There are two combinations of 0 and 5, and their joint probability is 0.00488, so it is 2* that or 0.00977
There are two combinations of 1 and 4, and that is 0.0488 when multiplied out.
There are two combinations of 2 and 3, and that is 0.0977
The sum of those 3 is 0.1563, assuming I understood the problem correctly.
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