Dark P.

asked • 05/14/18

Probability again

I dont’ quite understand how to go at this one, I feel like ther’s Not enough info in the question!  Any help is greatly appreciated.  According to the U.S. National Center for Health Statistics, the percentage of births done by Caesarean section in the United States was 22.8% in 1993. Based on this data, for five randomly chosen births in the United States, estimate the probability that
a) all five of the births were Caesareans.
b) none of the births were Caesareans.
c) At least one of the births were Ceasareans

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Kenneth S. answered • 05/15/18

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This should be approached as a BINOMIAL DISTRIBUTION (Bernoulli trials) probability question with n=5 and probability of success = 0.228
 
 
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Dark P.

I have absolutely no idea what that means lol.  
 
Can you show me the formula for that?
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05/16/18

Kenneth S.

To learn about this subject, I would suggest that you Google it. (Since it is advanced and has not yet been covered in your classes; you need not only the formula, but the reasoning behind it.)
 
a) P(5 consecutive Caesarean births) = 0.2285
b) P(none of 5 are Caesarean) = 0.7725
c) P(at least one is Caesarean) = 1-P(none) 
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05/16/18

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