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Conditional Probability of Baby Survival

  • Thread starter flemmyd
  • Start date
  • #1
144
1

Homework Statement


99% of all babies survive delivery. However, 10 % of all births involve Cesarean (C) sections, and when a C section is performed the baby survives 98% of the time. If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives?

Homework Equations


P(E) = 1 - P(EC)
P(E U F) = P(E)+P(F)-P(EF)
P(E|F)=P(EF)/P(F)

The Attempt at a Solution


P(S) = .99 for baby survival
P(C) = .1 for C section
P(S|C) = .98 for babies surviving the C section
P(S|CC) = ??

So I tried P(S|CC) =P(SCC)/P(CC)
I know the dominator = .9. I tried solving for the numerator using:
P(SCC)= P(S)+P(C)-P(S U C), but at this point, I don't know what the P(S U C) is.
 

Answers and Replies

  • #2
144
1
thanks to everyone who read/tried to help out, but i figured out the answer.

I just needed Bayes Theorem

P(E|C)*P(C)=P(C|B)P(B)
 

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