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hippie mii
Sep 20, 2007, 04:01 PM
98% of all babies survive delivery. However, 15% of all births involve Cesarean (C) sections, and when a C section is performed the baby survives 96% of the time.
If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives??

please help I cannot figure this out.

galactus
Sep 20, 2007, 04:59 PM
You can assume a number and fill in a chart. Makes these problems much easier.
Assume 1000 pregnat women.




SURVIVE NOT SURVIVE
-----------------------------------------------------------------------
| | |
CAESAREAN | 144 6 150
| | |
| | |
-----------------------------------------------------------------------
| |
| 836 | 14 | 850
NO CAESAREAN | |
| | |

980 20 1000


The probability a baby survives given the woman did not have a C-section is ?

I hope you can understand the chart. It got rather discombobulated.

hippie mii
Sep 20, 2007, 05:26 PM
I believe this chart says that the intersection of the event that the baby survives and the event that the woman does not have a C section is 836/1000 or .836
I think to get the answer to the original question "Given the woman does not have a c section, what is the probability that her baby survives?" you must take the intersection the two events above which is .836 and divide by the probability that the woman does not have a C section which is .85
this should give the final answer which is .9835