Q3.16 - A First Course in Probability

Question

Ninety-eight percent of all babies survive delivery. However, 15 percent of all births involve Cesarean (C) sections, and when a C section is performed, the baby survives 96 percent of the time when a C section is performed, the baby survives 96 percent of the time . If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives? 

Step-by-Step Solution

Answer

 The probability that a randomly chosen pregnant woman does not have a C section, is that her baby survives  is 0.9835

Step 1 Given Information

Given that ninety-eight percent of all babies survive delivery. 

15 percent of all births involve Cesarean (C) sections, 

when a C section is performed, the baby survives 96 percent of the time 

We have to find the probability that a baby survives  at randomly chosen pregnant women does not have a C section,  

Step 2 Explanation

Diagram

Decision Tree Analysis

Step 3: Explanation of Diagram

C : Cesarian delivery,$C^C$ : Normal delivery,$S$: Survival,D: Death

$P( \mathrm{S})=0.15\times 0.96+0.85\times x$

P(S)=0.98 given

$0.98=0.15x0.96+0.85x x$

$\Rightarrow x=0.9835$

Ste 4 Final Answer

 The probability that a randomly chosen pregnant woman does not have a C section, is that her baby survives  is 0.9835