Question 1166510
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The probability that at least one student tests positive is equal to 1 minus the probability that no student tests positive.  In general, the probability of any given number of positives could NOT be calculated using the binomial distribution because the events are not independent, i.e. the probability that student A tests positive is dependent on the disease status of the students that interact with that student.  But since we are actually calculating the probability that no one is positive, we can get away with using the stated positivity rate.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(=r,n,p)\ =\ {{n}\choose{r}}\,(p)^r\,(1-p)^{n-r}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(0,2000,0.0004)\ =\ {{2000}\choose{0}}\,\(0.0004\)^0\,\cdot\,\(0.9996\)^{2000}\ =\ 1\,\cdot\,1\,\cdot\(0.9996\)^{2000}\ \approx\ 0.45]


So the probability of at least one is approximately *[tex \Large 1\ -\ 0.45\ =\ 0.55]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish]


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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