Question 1178875
<font face="Times New Roman" size="+2">


Your answer to part a) is spot on.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {{19}\choose{19}}\cdot\(0.5)^{19}(0.5)^0]


Which reduces to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\cdot\(0.5)^{19}\cdot\,1\ \approx\ 1.907E-6]


as you reported.


Since your set up said "about 3 million" births in 4800 hospitals, I'm going to make it 3,009,600 births in 4800 hospitals which means 627 births per hospital and 33 sets of 19 births per hospital.


So your part b wants the probability of one hospital, i.e. 33 sets of 19 births, to have a string of 19 boys out of 33 tries.  The thing is, any number of sets of 19 boys from 1 to 33 would qualify. So what you want is the probability that in any one hospital what is the probability of <i>at least</i> one set of 19 boys.  Straight up, this is a seriously ugly mess of arithmetic because you need to compute, where *[tex \Large p\ =\ 1.907E-6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{r=1}^{33}\,{{33}\choose{r}}\,(p)^r(1-p)^{33-r}]


Which would require computing 33 terms and then summing them all. However, we can simplify the computation considerably by considering the fact that the probability of something happening at least once is equal to 1 minus the probability that it doesn't happen at all.  For this situation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ P(\geq{1},33,p)\ =\ 1\ -\ {{33}\choose{0}}\cdot(p)^0(1-p)^{33}\ =\ 1\cdot\ 1\cdot (1-p)^{33}]


For part c), I'll refer to the result of part b as *[tex \Large p_1]


Again we are faced with an "at least" situation, so we need to compute 1 minus the probability that it doesn't happen at all, in other words, 1 minus the probability of zero successes in 4800 trials where the probability of success on any given trial is *[tex \Large p_1]



*[tex \LARGE \ \ \ \ \ \ \ \ \ P(\geq{1},4800,p_1)\ =\ 1\ -\ {{4800}\choose{0}}\cdot(p_1)^0(1-p_1)^{4800}\ =\ 1\cdot\ 1\cdot (1-p_1)^{4800}]


You can do your own arithmetic.

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>