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?

Short answer

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

Step-by-step solution

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,

Explanation

Diagram

Decision Tree Analysis

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

$0.98=0.15x0.96+0.85xx$

$\Rightarrow x=0.9835$

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